Class 11 - Maths - Straight Lines

Exercise 10.1

Question 1:

Draw a quadrilateral in the Cartesian plane, whose vertices are (–4, 5), (0, 7), (5, –5) and          (-4,-2). Also, find its area.

Answer 1:

Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5) and D (-4, -2).

Then, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA, the

given quadrilateral can be drawn as

        Class_11_Straight_Lines_Graph1                         

To find the area of quadrilateral ABCD, we draw one diagonal AC.

Accordingly, area (ABCD) = area (∆ABC) + area (∆ACD)

We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is

(1/2)|x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)|

Therefore, area of ∆ABC = (1/2)|-4(7 + 5) + 0(-5 - 5) + 5(5 - 7)|

                                             = (1/2)|-4 * 12 – 5 * 2|

                                             = (1/2)|-48 – 10|

                                             = (1/2)|-58|

                                             = 58/2

                                             = 29 unit2       

Area of ∆ACD = (1/2)|-4(-5 + 2) + 5(-2 - 5) + (-4)(5 + 5)|

                         = (1/2)|4 * 3 – 5 * 7 – 4 * 10|

                         = (1/2)|12 -35 – 40|

                         = (1/2)|-63|

                         = 63/2  unit2       

Thus, area(ABCD) = 29 + 63/2 = (58 + 63)/2 = 121/2 unit2

Question 2:

The base of an equilateral triangle with side 2a lies along they y-axis such that the mid-point of the base is at the origin.

Find vertices of the triangle.

Answer:

Let ABC be the given equilateral triangle with side 2a.

So, AB = BC = CA = 2a

Class_11_Straight_Lines_Graph2

Let the base BC lies along the y-axis such that the mid-point of BC is at the origin.

i.e., BO = OC = a, where O is the origin.

Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are

(0, -a).

It is known that the line joining a vertex of an equilateral triangle with the mid-point of its

opposite side is perpendicular.

Hence, vertex A lies on the y-axis.

On applying Pythagoras theorem to ∆AOC, we obtain

     (AC)2 = (OA)2 + (OC)2

=> (2a)2 = (OA)2 + a2

=> 4a2 – a2 = (OA)2

=> (OA)2 = 3a2

=> OA = ±a√3

So, the coordinates of point A = (±a√3, 0)

Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and (a√3, 0) or (0, a),  

(0, -a), and (-a√3, 0).

Question 3:

Find the distance between P(x1, y1) and Q(x2, y2) when:

(i) PQ is parallel to the y-axis,  (ii) PQ is parallel to the x-axis.

Answer:

The given points are P(x1, y1) and Q(x2, y2).

(i) When PQ is parallel to the y-axis then x1 = x2

In this case, distance between P and Q = √{(x1 – x1)2 + (y2 – y1)2}

                                                                      = √{(y2 – y1)2}

                                                                      = |y2 – y1|       

(ii)When PQ is parallel to the x-axis then y1 = y2

In this case, distance between P and Q = √{(x2 – x1)2 + (y1 – y1)2}

                                                                      = √{(x2 – x1)2}

                                                                      = |x2 – x1|     

Question 4:

Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Answer:

Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).

Now, √{(7 – a)2 + (6 – 0)2} = √{(3 – a)2 + (4 – 0)2}

=> √(49 + a2 – 14a + 36) = √(9 + a2 – 6a + 16)

=> √(a2 – 14a + 85) = √(a2 – 6a + 25)

On squaring both sides, we get

=> a2 – 14a + 85 = a2 – 6a + 25

=> -14a + 6a = 25 – 85

=> -8a = -60

=> a = 60/8

=> a = 15/2

Thus, the required point on the x-axis is (15/2, 0)

Question 5:

Find the slope of a line, which passes through the origin and the mid-point of the line segment joining the points P (0, –4) and B (8, 0).

Answer:

The coordinates of the mid-point of the line segment joining the points P (0, –4) and B (8, 0)

are {(0 + 8)/2, (-4 + 0)/2} = (4, -2)

It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and

(x2, y2) is given by m = (y2 – y1)/(x2 – x1), x2 ≠ x1      

Therefore, the slope of the line passing through (0, 0) and (4, –2) is

(-2 – 0)/(4 – 0) = -2/4 = -1/2

Hence, the required slope of the line is -1/2

 

Question 6:

Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.

Answer:

The vertices of the given triangle are A (4, 4), B (3, 5) and C (–1, –1).

It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and

(x2, y2) is given by m = (y2 – y1)/(x2 – x1), x2 ≠ x1      

Slope of AB (m1) = (5 – 4)/(3 – 4) = -1

Slope of BC (m2) = (-1 – 5)/(-1 – 3) = -6/-4 = 3/2

Slope of CA (m3) = (4 + 1)/(4 + 1) = 5/5 = 1

It is observed that m1 * m3 = –1

This shows that line segments AB and CA are perpendicular to each other i.e., the given

triangle is right-angled at A (4, 4).

Thus, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.

Question 7:

Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

Answer:

If a line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise,

then the angle made by the line with the positive direction of the x-axis measured anti –clockwise is 90° + 30° = 120°.

Thus, the slope of the given line is

tan 120° = tan(180° – 60°) = –tan 60° = -√3

 Class_11_Straight_Lines_Graph3

Question 8:

Find the value of x for which the points (x, –1), (2, 1) and (4, 5) are collinear.

Answer:

If points A (x, –1), B (2, 1), and C (4, 5) are collinear, then

      Slope of AB = Slope of BC

=> {1 – (-1)}/(2 - x) = (5 - 1)/(4 - 2)

=> (1 + 1)/(2 - x) = 4/2

=> 2/(2 - x) = 2

=> 2 = 2(2 – x)

=> 2 = 4 – 2x

=> 2x = 4 – 2

=> 2x = 2

=> x = 1

Thus, the required value of x is 1

Question 9:

Without using distance formula, show that points (–2, –1), (4, 0), (3, 3) and (–3, 2) are vertices of a parallelogram.

Answer:

Let points (–2, –1), (4, 0), (3, 3), and (–3, 2) be respectively denoted by A, B, C and D.

Class_11_Straight_Lines_Parallelogram

Slope of AB = (0 + 1)/(4 + 2) = 1/6

Slope of CD = (2 - 3)/(-3 - 3) = -1/(-6) = 1/6

Since Slope of AB = Slope of CD

=> AB and CD are parallel to each other.

Now, slope of BC = (3 - 0)/(3 - 4) = 3/(-1) = -3

Slope of AD = (2 + 1)/(-3 + 2) = 3/(-1) = -3

=> Slope of BC = Slope of AD

=> BC and AD are parallel to each other.

Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD is a

parallelogram.

Thus, the points (–2, –1), (4, 0), (3, 3), and (–3, 2) are the vertices of a parallelogram.

Question 10:

Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).

Answer:

The slope of the line joining the points (3, –1) and (4, –2) is

m = {-2 – (-1)}/(4 - 3) = -2 + 1 = -1

Now, the inclination (θ) of the line joining the points (3, –1) and (4, – 2) is given by

      tan θ = –1

=> tan θ = tan 135°

⇒ θ = 135°

Thus, the angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°.

Question 11:

The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.

Answer:

Let m1 and m be the slopes of the two given lines such that m1 = 2m.

We know that if θ is the angle between the lines l1 and l2 with slopes m1 and m2, then

tan θ = |(m2 – m1)/(1 + m1 m2)|

It is given that the tangent of the angle between the two lines is 1/3

=> 1/3 = |(m – 2m)/(1 + 2m * m)|

=> 1/3 = |-m/(1 + 2m2)|

=> 1/3 = -m/(1 + 2m2) or 1/3 = -{-m/(1 + 2m2)}

Case I:

      1/3 = -m/(1 + 2m2)

=> 1 + 2m2 = -3m

=> 2m2 + 3m + 1 = 0

=> (m + 1)(2m + 1) = 0

=> m = -1, -1/2

If m = –1, then the slopes of the lines are –1 and –2

If m = -1/2, then the slopes of the lines are -1/2 and –1

Case II:

      1/3 = -{-m/(1 + 2m2)}

=> 1/3 = m/(1 + 2m2)

=> 1 + 2m2 = 3m

=> 2m2 - 3m + 1 = 0

=> (m - 1)(2m - 1) = 0

=> m = 1, 1/2

If m = 1, then the slopes of the lines are 1 and 2.

If m = 1/2, then the slopes of the lines are 1/2 and 1

Hence, the slopes of the lines are –1 and –2 or -1/2 and –1 or 1 and 2 or 1/2 and 1.

Question 12:

A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k – y1 = m(h – x1).

Answer:

The slope of the line passing through (x1, y1) and (h, k) is (k – y1)/(h – x1)

It is given that the slope of the line is m.

=> m = (k – y1)/(h – x1)

=> k – y1 = m(h – x1)

Hence, k – y1 = m(h – x1)

Question 13:

If three point (h, 0), (a, b) and (0, k) lie on a line, show that a/h + b/k = 1

Answer:

If the points A (h, 0), B (a, b), and C (0, k) lie on a line, then

      Slope of AB = Slope of BC

=> (b - 0)/(a - h) = (k - b)/(0 - a)

=> b/(a - h) = (k - b)/(-a)

=> -ab = (k - b)(a - h)

=> -ab = ka – kh – ab + bh

=> ka + bh = kh

On dividing both sides by kh, we get

=> ka/kh + bh/kh = kh/kh

=> a/h + b/k = 1

Hence, a/h + b/k = 1

Question 14:

Consider the given population and year graph. Find the slope of the line AB and using it,

find what will be the population in the year 2010?

 

Answer:

Since line AB passes through points A (1985, 92) and B (1995, 97), its slope is

(97 - 92)/(1995 - 1985) = 5/10 = 1/2

Let y be the population in the year 2010. Then, according to the given graph, line AB must pass

through point C (2010, y).

Class_11_Straight_Lines_Graph4

So, Slope of AB = Slope of BC

=> 1/2 = (y - 97)/(2010 - 1995)

=> 1/2 = (y - 97)/15

=> 15/2 = y – 97

=> y – 97 = 7.5

=> y = 97 + 7.5

=> y = 104.5

Thus, the slope of line AB is 1/2 while in the year 2010, the population will be 104.5 crores.

 

                                                                  Exercise 10.2

Question 1:

Write the equations for the x and y-axes.

Answer:

The y-coordinate of every point on the x-axis is 0.

Therefore, the equation of the x-axis is y = 0.

The x-coordinate of every point on the y-axis is 0.

Therefore, the equation of the y-axis is x = 0.

Question 2:

Find the equation of the line which passes through the point (–4, 3) with slope 1/2.

Answer:

We know that the equation of the line passing through point (x0, y0), whose slope is m, is

y – y0 = m(x – x0)

Thus, the equation of the line passing through point (–4, 3), whose slope is 1/2, is

       y – 3 = (x + 4)/2

=> 2y – 6 = x + 4

=> x + 4 – 2y + 6 = 0

=> x – 2y + 10 = 0

Question 3:

Find the equation of the line which passes though (0, 0) with slope m.

Answer:

We know that the equation of the line passing through point (x0, y0), whose slope is m, is

y – y0 = m(x – x0)

Thus, the equation of the line passing through point (0, 0), whose slope is m, is

      y – 0 = m(x – 0)

=> y = mx

Question 4:

Find the equation of the line which passes through (2, 2√3) and is inclined with the x-axis at an angle of 75°.

Answer:

The slope of the line that inclines with the x-axis at an angle of 75° is

      m = tan 75°

=> m = tan(45° + 30°)

=> m = (tan 45° + tan 30°)/(1 - tan 45° * tan 30°)

=> m = (1 + 1/√3)/(1 - 1 * 1/√3)

=> m = (1 + 1/√3)/(1 - 1/√3)

=> m = (√3 + 1)/(√3 - 1)

We know that the equation of the line passing through point (x0, y0), whose slope is m, is

y – y0 = m(x – x0)

Thus, if a line passes though (2, 2√3) and inclines with the x-axis at an angle of 75°, then the

equation of the line is given as

      (y - 2√3) = {(√3 + 1)/(√3 - 1)}(x - 2)

=> (y - 2√3)(√3 - 1) = (√3 + 1)(x - 2)

=> y(√3 - 1) - 2√3(√3 - 1) = x(√3 + 1) - 2(√3 + 1)

=> y(√3 - 1) - 6 + 2√3 = x(√3 + 1) - 2√3 – 2

=> x(√3 + 1) - 2√3 – 2 - y(√3 - 1) + 6 - 2√3 = 0

=> x(√3 + 1) - y(√3 - 1) + 4 - 4√3 = 0

=> x(√3 + 1) - y(√3 - 1) = 4(√3 - 1)

Question 5:

Find the equation of the line which intersects the x-axis at a distance of 3 units to the left of origin with slope –2.

Answer:

It is known that if a line with slope m makes x-intercept d, then the equation of the line is

given as y = m(x – d)

For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d = –3.

The slope of the line is given as m = –2

Thus, the required equation of the given line is

=> y = –2[x – (–3)]

=> y = –2[x + 3]

=> y = –2x – 6

=> 2x + y + 6 = 0

Question 6:

Find the equation of the line which intersects the y-axis at a distance of 2 units above the origin and makes an angle of 30°

with the positive direction of the x-axis.

Answer:

It is known that if a line with slope m makes y-intercept c, then the equation of the line

is given as y = mx + c

Here, c = 2 and m = tan 30° = 1/√3

Thus, the required equation of the given line is

=> y = x/√3 + 2

=> y = (x + 2√3)/ √3

=> y√3 = x + 2√3

=> x - √3y + 2√3 = 0

Question 7:

Find the equation of the line which passes through the points (–1, 1) and (2, –4).

Answer:

It is known that the equation of the line passing through points (x1, y1) and (x2, y2) is

(y – y1) = {(y2 – y1)/(x2 – x1)}(x – x1)

Therefore, the equation of the line passing through the points (–1, 1) and (2, –4) is

      (y – 1) = {(-4 – 1)/(2 + 1)}(x + 1)

=> (y – 1) = (-5/3)(x + 1)

=> 3(y – 1) = -5(x + 1)

=> 3y – 3 = -5x – 5

=> 5x + 3y + 2 = 0

Question 8:

Find the equation of the line which is at a perpendicular distance of 5 units from the origin

and the angle made by the perpendicular with the positive x-axis is 30°

Answer:

If p is the length of the normal from the origin to a line and ω is the angle made by the normal

with the positive direction of the x-axis, then the equation of the line is given by

x cos ω + y sin ω = p.

Here, p = 5 units and ω = 30°

Thus, the required equation of the given line is x

x cos 30° + y sin 30° = 5

     √3x/2 + y/2 = 5

=> √3x + y = 5 * 2

=> √3x + y = 10

Question 9:

The vertices of ∆PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.

Answer:

It is given that the vertices of ∆PQR are P (2, 1), Q (–2, 3), and R (4, 5).

Let RL be the median through vertex R.

Accordingly, L is the mid-point of PQ.

By mid-point formula, the coordinates of point L are given by {(2 - 2)/2, (1 + 3)/2} = (0, 2)

          Class_11_Straight_Lines_Triangle                                               

It is known that the equation of the line passing through points (x1, y1) and (x2, y2) is

(y – y1) = {(y2 – y1)/(x2 – x1)}(x – x1)

Therefore, the equation of RL can be determined by substituting

(x1, y1) = (4, 5) and (x2, y2) = (0, 2)

      (y – 5) = {(2 – 5)/(0 - 4)}(x - 4)

=> (y – 5) = (-3/-4)(x - 4)

=> 4(y – 5) = 3(x - 4)

=> 4y – 20 = 3x – 12

=> 3x - 4y + 8 = 0

Thus, the required equation of the median through vertex R is 3x – 4y + 8 = 0

Question 10:

Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

Answer:

The slope of the line joining the points (2, 5) and (–3, 6) is

Class_11_Straight_Lines_LineSegment

m = (6 - 5)/(-3 - 2) = -1/5

We know that two non-vertical lines are perpendicular to each other if and only if their slopes

are negative reciprocals of each other.

Therefore, slope of the line perpendicular to the line through the points (2, 5) and (–3, 6)

= -1/m = -1/(-1/5) = 5

Now, the equation of the line passing through point (–3, 5), whose slope is 5, is

      y - 5 = 5(x + 3)

=> y - 5 = 5x + 15

=> 5x – y + 20 = 0

Question 11:

A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1 : n.

Find the equation of the line.

Answer:

According to the section formula, the coordinates of the point that divides the line segment

joining the points (1, 0) and (2, 3) in the ratio 1 : n is given by

[(n * 1 + 1 * 2)/(1 + n), (n * 0 + 1 * 3)/(1 + n)] = [(n + 2)/(n + 1), 3/(n + 1)]

The slope of the line joining the points (1, 0) and (2, 3) is

m = (3 - 0)/(2 - 1) = 3

We know that two non-vertical lines are perpendicular to each other if and only if their slopes

are negative reciprocals of each other.

Therefore, slope of the line that is perpendicular to the line joining the points (1, 0) and (2, 3)

= -1/m = -1/3

Now, the equation of the line passing through [(n + 2)/(n + 1), 3/(n + 1)] and whose slope is

-1/3 given by

      y - 3/(n + 1) = (-1/3)[x – (n + 2)/(n + 1)]

=> y(n + 1) - 3 = (-1/3)[x(n + 1) – (n + 2)]

=> 3y(n + 1) - 9 = -[x(n + 1) – (n + 2)]

=> 3y(n + 1) - 9 + x(n + 1) – (n + 2) = 0

=> (n + 1)x - 3(n + 1)y – 9 – n – 2 = 0

=> (n + 1)x - 3(n + 1)y – n – 11 = 0

=> (n + 1)x - 3(n + 1)y = n + 11

Question 12:

Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Answer:

The equation of a line in the intercept form is

x/a + y/b = 1   …………..1

Here, a and b are the intercepts on x and y axes respectively.

It is given that the line cuts off equal intercepts on both the axes.

This means that a = b.

Now equation 1 becomes

      x/a + y/a = 1

=> x + y = a       ……..2

Since the given line passes through point (2, 3), equation 2 reduces to

      2 + 3 = a

=> a = 5

On substituting the value of a in equation 2, we get

x + y = 5

Which is the required equation of the line.

Question 13:

Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Answer:

The equation of a line in the intercept form is

x/a + y/b = 1   ………1

Here, a and b are the intercepts on x and y axes respectively.

It is given that

      a + b = 9

=> b = 9 – a     ............2

From equations 1 and 2, we obtain

x/a + y/(9 - a) = 1  ………..3

It is given that the line passes through point (2, 2). Therefore, equation 3 becomes

      2/a + 2/(9 - a) = 1

=> 2{1/a + 1/(9 - a)} = 1

=> 2[(9 - a + a)/{a(9 - a)}] = 1

=> 18/{9a – a2} = 1

=> 18 = 9a – a2

=> a2 – 9a + 18 = 0

=> (a - 6)(a - 3) = 0

=> a = 3, 6

If a = 6 and b = 9 – 6 = 3, then the equation of the line is

      x/6 + y/3 = 1

=> (x + 2y)/6 = 1

=> x + 2y = 6

If a = 3 and b = 9 – 3 = 6, then the equation of the line is

      x/3 + y/6 = 1

=> (2x + y)/6 = 1

=> 2x + y = 6

Question 14:

Find equation of the line through the point (0, 2) making an angle 2π/3 with the positive x-axis.

Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Answer:

The slope of the line making an angle 2π/3 with the positive x-axis is

m = tan 2π/3 = -√3

Now, the equation of the line passing through point (0, 2) and having a slope -√3 is

      (y - 2) = -√3(x - 0)

=> y - 2 = -√3x

=> √3x + y – 2 = 0

The slope of line parallel to √3x + y – 2 = 0 is -√3

It is given that the line parallel to line √3x + y – 2 = 0 crosses the y-axis 2 units below the origin

i.e., it passes through point (0, –2).

Hence, the equation of the line passing through point (0, –2) and having a slope is

      y – (-2) = -√3(x – 0)

=> y + 2 = -√3x

=> √3x + y + 2 = 0

Question 15:

The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

Answer:

The slope of the line joining the origin (0, 0) and point (–2, 9) is

m1 = (9 - 0)/(-2 - 0) = -9/2

Again, the slope of the line perpendicular to the line joining the origin and point (– 2, 9) is

m2 = -1/ m1 = -1/(-9/2) = 2/9

Now, the equation of the line passing through point (–2, 9) and having a slope m2 is

       y – 9 = (2/9)(x + 2)

=> 9(y – 9) = 2(x + 2)

=> 9y – 81 = 2x + 4

=> 2x – 9y + 85 = 0

Question 16:

The length L (in centimeter) of a copper rod is a linear function of its Celsius temperature C.

In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.

Answer:

It is given that when C = 20, the value of L is 124.942

and when C = 110, the value of L is 125.134

Now, points (20, 124.942) and (110, 125.134) satisfy the linear relation between L and C.

Assuming C along the x-axis and L along the y-axis, we have two points i.e., (20, 124.942) and

(110, 125.134) in the XY plane.

Therefore, the linear relation between L and C is the equation of the line passing through

points (20, 124.942) and (110, 125.134)

      L – 124.942 = {(125.134 - 124.942)/(110 - 20)}(C - 20)

=> L – 124.942 = (0.192/90)(C - 20)

=> L = (0.192/90)(C - 20) + 124.942

Which is the required linear relation.

Question 17:

The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre

and 1220 litres of milk each week at Rs 16/litre.

Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?

Answer:

The relationship between selling price and demand is linear.

Assuming selling price per litre along the x-axis and demand along the y-axis, we have

two points i.e., (14, 980) and (16, 1220) in the XY plane that satisfy the linear relationship

between selling price and demand.

Therefore, the linear relationship between selling price per litre and demand is the equation

of the line passing through points (14, 980) and (16, 1220).

      y – 980 = {(1220 - 980)/(16 - 14)}(x - 14)

=> y – 980 = (240/2)(x - 14)

=> y – 980 = 120(x - 14)

=> y = 120(x - 14) + 980

When x = Rs 17/litre,

=> y = 120(17 - 14) + 980

=> y = 120 * 3 + 980

=> y = 360 + 980

=> y = 1340

Thus, the owner of the milk store could sell 1340 litres of milk weekly at Rs 17/litre.

Question 18:

P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is

x/a + y/b = 2

Answer:

Let AB be the line segment between the axes and let P (a, b) be its mid-point.

Let the coordinates of A and B be (0, y) and (x, 0) respectively.

Since P (a, b) is the mid-point of AB,

      {(0 + x)/2, (0 + y)/2} = (a, b)

=> (x/2, y/2) = (a, b)

=> x/2 = a and y/2 = b

=> x = 2a and y = 2b

Thus, the respective coordinates of A and B are (0, 2b) and (2a, 0).

The equation of the line passing through points (0, 2b) and (2a, 0)

      y – 2b = {(0 – 2b)/(2a - 0)}(x - 0)

=> y – 2b = (-2b/2a)(x)

=> y – 2b = (-b/a)(x)

=> a(y – 2b) = -bx

=> ay – 2ab = -bx

=> bx + ay = 2ab

On dividing both sides by ab, we obtain

=> bx/ab + ay/ab = 2ab/ab

=> x/a + y/b = 2

Thus, the equation of the line is x/a + y/b = 2

Question 19:

Point R (h, k) divides a line segment between the axes in the ratio 1 : 2. Find equation of the line.

Answer:

Let AB be the line segment between the axes such that point R (h, k) divides AB in the ratio        

1: 2.

  Class_11_Straight_Lines_LineSegment1                                                         

Let the respective coordinates of A and B be (x, 0) and (0, y).

Since point R (h, k) divides AB in the ratio 1: 2, according to the section formula,

      (h, k) = {(1 * 0 + 2 * x)/(1 + 2), (1 * y + 2 * 0)/(1 + 2)}

=> (h, k) = (2x/3, y/3)

=> h = 2x/3 and k = y/3

=> x = 3h/2 and y = 3k

Therefore, the respective coordinates of A and B are (3h/2, 0) and (0, 3k).

Now, the equation of line AB passing through points (3h/2, 0) and (0, 3k) is

      y – 0 = {(3k - 0)/(0 – 3h/2)}(x – 3h/2)

=> y = (-2k/h)(x – 3h/2)

=> hy = (-2k)(x – 3h/2)

=> hy = (-2k)(x – 3h/2)

=> hy = -2kx + 3hk

=> 2kx + hy = 3kh

Thus, the required equation of the line is 2kx + hy = 3hk

Question 20:

By using the concept of equation of a line, prove that the three points (3, 0), (–2, –2) and (8, 2) are collinear.

Answer:

In order to show that points (3, 0), (–2, –2), and (8, 2) are collinear, it suffices to show that the

line passing through points (3, 0) and (–2, –2) also passes through point (8, 2).

The equation of the line passing through points (3, 0) and (–2, –2) is

      y – 0 = {(-2 – 0)/(-2 - 3)}(x - 3)

=> y = (-2/-5)(x - 3)

=> y = (2/5)(x - 3)

=> 5y = 2(x - 3)

=> 5y = 2x – 6

=> 2x – 5y = 6

It is observed that at x = 8 and y = 2,

L.H.S. = 2 * 8 – 5 * 2 = 16 – 10 = 6 = R.H.S.

Therefore, the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2).

Hence, the points (3, 0), (–2, –2), and (8, 2) are collinear.

 

                                                                     Exercise 10.3

Question 1:

Reduce the following equations into slope-intercept form and find their slopes and the y intercepts.                                                                                                              (i) x + 7y = 0                               (ii) 6x + 3y – 5 = 0                        (iii) y = 0

Answer:

(i) The given equation is x + 7y = 0. It can be written as

y = -x/7 + 0  ………..1

This equation is of the form y = mx + c, where m = -1/7 and c = 0

Therefore, equation 1 is in the slope-intercept form, where the slope and the y-intercept are

-1/7 and 0 respectively.

(ii) The given equation is 6x + 3y – 5 = 0.

It can be written as

      3y = -6x + 5

=> y = -6x/3 + 5/3

=> y = -2x + 5/3  ………….2

This equation is of the form y = mx + c, where m = -2 and c = 5/3

Therefore, equation 2 is in the slope-intercept form, where the slope and the y-intercept are

-2 and 5/3 respectively.

(iii) The given equation is y = 0.

It can be written as

y = 0 * x + 0   ……….3

This equation is of the form y = mx + c, where m = 0 and c = 0.

Therefore, equation 3 is in the slope-intercept form, where the slope and the y-intercept

are 0 and 0 respectively.

Question 2:

Reduce the following equations into intercept form and find their intercepts on the axes.

(i) 3x + 2y – 12 = 0                        (ii) 4x – 3y = 6                      (iii) 3y + 2 = 0

Answer:

(i) The given equation is 3x + 2y – 12 = 0

It can be written as

      3x + 2y = 12

=> 3x/12 + 2y/12 = 1

=> x/4 + y/6 = 1 …………1

This equation is in the form x/a + y/b = 1

Where a = 4 and b = 6

Therefore, equation 1 is in the intercept form, where the intercepts on the x and y axes are 4

and 6 respectively.

(ii) The given equation is 4x – 3y = 6

It can be written as

      4x/6 – 3y/6 = 1

=> 2x/3 – y/2 = 1

=> x/(3/2) + y/(-2) = 1   …………..2

This equation is of the form x/a + y/b = 1, where a = 3/2 and b = –2.

Therefore, equation 2 is in the intercept form, where the intercepts on the x and y axes are

3/2 and –2 respectively.

(iii) The given equation is 3y + 2 = 0

It can be written as

      3y = -2

=> 3y/(-2) = 1

=> y/(-2/3) = 1       …………..3

The equation is of the form x/a + y/b, where a = 0 and b = -2/3

Therefore, equation 3 is in the intercept form, where the intercept on the y-axis is -2/3 and it

has no intercept on the x-axis.

Question 3:

Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

(i) x – √3y + 8 = 0                        (ii) y – 2 = 0                     (iii) x – y = 4

Answer 3:

(i) The given equation is x – √3y + 8 = 0                       

It can be reduced as:

      x – √3y = -8

=> -x + √3y = 8                        

On dividing both sides by √{(-1)2 +(√3)2} = √(1 + 3) = √4 = 2, we get

=> -x/2 + √3y/2 = 8/2

=> -x/2 + √3y/2 = 4

=> (-1/2)x + (√3/2)y = 4

=> x cos 120° + y sin 120° = 4  ………….1

Equation 1 is in the normal form.

On comparing equation 1 with the normal form of equation of line x cos ω + y sin ω = p, we get

ω = 120° and p = 4

Thus, the perpendicular distance of the line from the origin is 4, while the angle between the

perpendicular and the positive x-axis is 120°.

(ii) The given equation is y – 2 = 0

It can be reduced as 0.x + 1.y = 2

On dividing both sides by √{02 + 12} = 1, we get

      0 * x + 1 * y = 2

=> x cos 90° + y sin 90° = 2  ………..1

Equation 1 is in the normal form.

On comparing equation 1 with the normal form of equation of line x cos ω + y sin ω = p, we get

ω = 90° and p = 2

Thus, the perpendicular distance of the line from the origin is 2, while the angle between the

perpendicular and the positive x-axis is 90°.

(iii) The given equation is x – y = 4

It can be reduced as 1 * x + (–1)y = 4

On dividing both sides by √{12 + (-1)2} = √2, we get

     (1/√2)x + (-1/√2) = 4/√2

=> x cos (2π – π/4) + y sin (2π – π/4) = 2√2

=> x cos 315° + y sin 315° = 2√2  ………..1

Equation (1) is in the normal form.

On comparing equation 1 with the normal form of equation of line x cos ω + y sin ω = p, we get

ω = 315° and p = 2√2

Thus, the perpendicular distance of the line from the origin is 2√2, while the angle between

the perpendicular and the positive x-axis is 315°.

Question 4:

Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

Answer:

The given equation of the line is 12(x + 6) = 5(y – 2)

=> 12x + 72 = 5y – 10

=> 12x – 5y + 82 = 0    ….....1

On comparing equation 1 with general equation of line Ax + By + C = 0, we get

A = 12, B = –5, and C = 82

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is

given by d = |Ax1 + By1 + C|/√(A2 + B2)

The given point is (x1, y1) = (–1, 1)

Therefore, the distance of point (–1, 1) from the given line

= |12 * (-1) + (-5) * 1 + 82|/√{(12)2 + (-5)2}

= |-12 - 5 + 82|/√ (144 + 25)

= |65|/√169

= 65/13

= 5 units

Question 5:

Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.

Answer:

The given equation of line is x/3 + y/4 = 1

=> (4x + 3y)/12 = 1

=> 4x + 3y = 12   ……………1

On comparing equation 1 with general equation of line Ax + By + C = 0, we get

A = 4, B = 3, and C = –12

Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is

given by d = |Ax1 + By1 + C|/√(A2 + B2)

Therefore,

      4 = |4a + 3 * 0 - 12|/√{42 + 32}

=> 4 = |4a - 12|/√ (16 + 9)

=> 4 = |4a – 12|/5

=> |4a – 12| = 20

=> 4a – 12 = 20 or –(4a - 12) = 20

=> 4a = 20 + 12 or 4a = -20 + 12

=> 4a = 32 or 4a = -8

=> a = 8 or a = -2

Thus, the required points on the x-axis are (–2, 0) and (8, 0).

Question 6:

Find the distance between parallel lines

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

(ii) l(x + y) + p = 0 and l(x + y) – r = 0

Answer:

It is known that the distance (d) between parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is

given by d = |C1 – C2|/√(A2 + B2)

 

(i) The given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.

Here, A = 15, B = 8, C1 = –34, and C2 = 31

Therefore, the distance between the parallel lines is

d = |C1 – C2|/√(A2 + B2) = |-34 – 31|/√(152 + 82) = |-65|/√ (225 + 64) = 65/17 units

(ii)The given parallel lines are l(x + y) + p = 0 and l(x + y) – r = 0

lx + ly + p = 0 and lx + ly – r = 0 Here,

A = l, B = l, C1 = p, and C2 = –r

Therefore, the distance between the parallel lines is

d = |C1 – C2|/√(A2 + B2) = |p + r|/√(l2 + l2) = |p + r|/√(2l2) = (1/√2) *|(p + r)/l| units

Question 7:

Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point  (–2, 3).

Answer:

The equation of the given line is 3x – 4y + 2 = 0

=> 4y = 3x + 2

=> y = 3x/4 + 2/4

=> y = 3x/4 + 1/2

which is of the form y = mx + c

So, slope of the given line = 3/4

It is known that parallel lines have the same slope.

Slope of the other line = m = 3/4

Now, the equation of the line that has a slope of 3/4 and passes through the point (–2, 3) is

      (y – 3) = 3{x – (-2)}/4

=> y – 3 = 3(x + 2)/4

=> 4(y – 3) = 3(x + 2)

=> 4y – 12 = 3x + 6

=> 3x – 4y + 18 = 0

Question 8:

Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Answer:

The given equation of line is x – 7y + 5 = 0

=> 7y = x + 5

=> y = x/7 + 5/7

which is of the form y = mx + c

So, the slope of the given line = 1/7

The slope of the line perpendicular to the line having a slope of 1/7 is

m = -1/(1/7) = -7

The equation of the line with slope –7 and x-intercept 3 is given by

     y = m (x – d)

=> y = –7(x – 3)

=> y = –7x + 21

=> 7x + y = 21

Question 9:

Find angles between the lines √3x + y = 1 and x + √3y = 1

Answer:

The given lines are √3x + y = 1 and x + √3y = 1

=> y = -√3x + 1  ………..1

and y = -x/√3 + 1/√3   ……..2

The slope of the 1 is m1 = -√3 and the slope of the line 2 is m2 = -1/√3

The acute angle i.e., θ between the two lines is given by

      tan θ = |(m1 – m2)/(1 + m1 * m2)|

=> tan θ = |(-√3 + 1/√3)/{1 + (-√3) * (-1/√3)}|

=> tan θ = |(-√3 + 1/√3)/(1 + 1)|

=> tan θ = |{(-3 + 1)/√3}/2|

=> tan θ = |-2/2√3|

=> tan θ = |-1/√3|

=> tan θ = 1/√3

=> tan θ = tan 30°

=> θ = 30°

Thus, the angle between the given lines is either 30° or 180° – 30° = 150°.

Question 10:

The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0, at right angle. Find the value of h.

Answer:

The slope of the line passing through points (h, 3) and (4, 1) is

m1 = (1 - 3)/(4 - h) = -2/(4 - h)

Given line is 7x – 9y – 19 = 0

=> 9y = 7x – 19

=> y = 7x/9 – 19/7

The slope of line 7x – 9y – 19 = 0 is m2 = 7/9

It is given that the two lines are perpendicular.

So, m1 * m2 = -1

=> {-2/(4 - h)} * (7/9) = -1

=> -14/(36 – 9h) = -1

=> 36 – 9h = 14

=> 9h = 36 – 14

=> 9h = 22

=> h = 22/9

Thus, the value of h is 22/9

Question 11:

Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is  A(x –x1) + B(y – y1) = 0.

Answer:

The slope of line Ax + By + C = 0

=> y = (-A/B)x + (-C/B)

So, m = -A/B

It is known that parallel lines have the same slope.

So, the slope of the other line = m = -A/B

The equation of the line passing through point (x1, y1) and having a slope –A/B is

      y – y1 = m(x – x1)

=> y – y1 = (-A/B)(x – x1)

=> B(y – y1) = -A(x – x1)

=> A(x – x1) + B(y – y1) = 0

Hence, the line through point (x1, y1) and parallel to line Ax + By + C = 0 is

A(x –x1) + B(y – y1) = 0

Question 12:

Two lines passing through the point (2, 3) intersects each other at an angle of 60°.

If slope of one line is 2, find equation of the other line.

Answer:

It is given that the slope of the first line, m1 = 2

Let the slope of the other line be m2

The angle between the two lines is 60°.

      tan θ = |(m1 – m2)/(1 + m1 * m2)|

=> tan 60° = |(2 - m2)/(1 + 2m2)|

=> √3 = ±(2 - m2)/(1 + 2m2)

=> √3 = (2 - m2)/(1 + 2m2) or √3 = -(2 - m2)/(1 + 2m2)

=> √3(1 + 2m2) = (2 - m2) or √3(1 + 2m2) = -(2 - m2)

=> √3 + 2√3m2 = 2 - m2 or √3 + 2√3m2 = -2 + m2

=> √3 + 2√3m2 + m2 = 2 or √3 + 2√3m2 - m2 = -2

=> (2√3 + 1)m2 = 2 - √3 or (2√3 – 1)m2 = -2 - √3

=> m2 = (2 - √3)/ (2√3 + 1) or m2 = -(2 + √3)/(2√3 - 1)

Case 1: When m2 = (2 - √3)/(2√3 + 1)

The equation of the line passing through point (2, 3) and having a slope of

m2 = (2 - √3)/(2√3 + 1) is given by

      (y – 3) = {(2 - √3)/ (2√3 + 1)}(x – 2)

=> (y – 3)(2√3 + 1) = (2 - √3)(x – 2)

=> (2√3 + 1)y - 3(2√3 + 1) = (2 - √3)x – 2(2 - √3)

=> (√3 - 2)x + (2√3 + 1)y = -4 + 2√3 + 6√3 + 3

=> (√3 - 2)x + (2√3 + 1)y = -1 + 8√3

In this case, the equation of the other line is (√3 - 2)x + (2√3 + 1)y = -1 + 8√3

Case 2: When m2 = -(2 + √3)/ (2√3 - 1)

The equation of the line passing through point (2, 3) and having a slope of

m2 = -(2 + √3)/(2√3 + 1) is given by

      (y – 3) = {-(2 + √3)/ (2√3 - 1)}(x – 2)

=> (y – 3)(2√3 - 1) = -(2 + √3)(x – 2)

=> (2√3 - 1)y - 3(2√3 - 1) = -(2 + √3)x + 2(2 + √3)

=> (2 + √3)x + (2√3 - 1)y = 4 + 2√3 + 6√3 - 3

=> (2 + √3)x + (2√3 - 1)y = 1 + 8√3

In this case, the equation of the other line is (2 + √3)x + (2√3 - 1)y = 1 + 8√3

Question 13:

Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).

Answer:

The right bisector of a line segment bisects the line segment at 90°.

The end-points of the line segment are given as A (3, 4) and B (–1, 2).

Mid-point of AB = {(3 - 1)/2, (4 + 2)/2} = (1, 3)

Slope of AB = (2 - 4)/(-1 - 3) = -2/-4 = 1/2

So, the slope of the line perpendicular to AB = -1/(1/2) = -2

The equation of the line passing through (1, 3) and having a slope of –2 is

     (y – 3) = –2(x – 1)

=> y – 3 = –2x + 2

=> 2x + y = 5

Thus, the required equation of the line is 2x + y = 5.

Question 14:

Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line  3x – 4y – 16 = 0.

Answer:

Let (a, b) be the coordinates of the foot of the perpendicular from the point (–1, 3) to the line

3x – 4y – 16 = 0.

Class_11_Straight_Lines_LineSegment2

Slope of the line joining (–1, 3) and (a, b), m1 = (b - 3)/(a + 1)

Given line is 3x – 4y – 16 = 0

=> 4y = 3x – 16

=> y = 3x/4 – 4

Slope of this line is m2 = 3/4

Since these two lines are perpendicular, m1 * m2 = –1

=> {(b - 3)/(a + 1)} * (3/4) = -1

=> (3b - 9)/(4a + 4) = -1

=> 3b – 9 = -4a – 4

=> 4a + 3b = 5   ………….1

Point (a, b) lies on line 3x – 4y = 16

3a – 4b = 16   ..........2

On solving equations 1 and 2, we obtain

A = 68/25 and b = -49/25

Thus, the required coordinates of the foot of the perpendicular is (68/25, -49/25).

Question 15:

The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c.

Answer:

The given equation of line is y = mx + c.

It is given that the perpendicular from the origin meets the given line at (–1, 2).

Therefore, the line joining the points (0, 0) and (–1, 2) is perpendicular to the given line.

Slope of the line joining (0, 0) and (–1, 2) = 2/(-1) = -2

The slope of the given line is m.

So, m * (-2) = -1

=> m = 1/2            [Since two lines are perpendicular]

Since point (–1, 2) lies on the given line, it satisfies the equation y = mx + c.

=> 2 = m(-1) + c

=> 2 = (1/2) * (-1) + c

=> 2 = -1/2 + c

=> c = 2 + 1/2

=> c = 5/2

Thus, the respective values of m and c are 1/2 and 5/2.

Question 16:

If p and q are the lengths of perpendiculars from the origin to the lines x cos θ – y sin θ = k cos 2θ and x sec θ + y cosec θ = k,

respectively, prove that p2 + 4q2 = k2

Answer:

The equations of given lines are

x cos θ – y sinθ = k cos 2θ ...................1

x sec θ + y cosec θ = k .........................2

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

d = |Ax1 + By1 + C|/√(A2 + B2)

On comparing equation 1 to the general equation of line i.e., Ax + By + C = 0, we get

A = cos θ, B = –sin θ, and C = –k cos 2θ

It is given that p is the length of the perpendicular from (0, 0) to line 1.

So, p = |A * 0 + B * 0 + C|/√(A2 + B2)

          = |C|/√(cos2 θ  + sin2 θ)

          = |-k cos 2θ|/√(cos2 θ  + sin2 θ)

          = |-k cos 2θ|      ………3

It is given that q is the length of the perpendicular from (0, 0) to line 2.

So, q = |A * 0 + B * 0 + C|/√(A2 + B2)

          = |C|/√(sec2 θ  + cosec2 θ)

          = |-k|/√(sec2 θ  + cosec2 θ)      …………..4

From equation 3 and 4, we have

p2 + 4q2 = (|-k cos 2θ|)2 + 4[|-k|/√(sec2 θ  + cosec2 θ)]2

               = k2 cos2 2θ + 4 k2/(sec2 θ  + cosec2 θ)

               = k2 cos2 2θ + 4 k2/(1/cos2 θ  + 1/sin2 θ)

               = k2 cos2 2θ + 4 k2/{(cos2 θ  + sin2 θ)/( cos2 θ  * sin2 θ)}      

               = k2 cos2 2θ + 4 k2/{1/( cos2 θ  * sin2 θ)}

               = k2 cos2 2θ + 4 k2 ( cos2 θ  * sin2 θ)

               = k2 cos2 2θ + k2 ( 4cos2 θ  * sin2 θ)

               = k2 cos2 2θ + k2 ( 2cos  θ  * sin θ) 2          

               = k2 cos2 2θ + k 2 sin 2

               = k2(cos2 2θ + sin 2 2θ)

              = k2          

Hence, we proved that p2 + 4q2 = k2

Question 17:

In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

Answer:

Let AD be the altitude of triangle ABC from vertex A.

Accordingly, AD Ʇ BC

Class_11_Straight_Lines_Triangle_1

The equation of the line passing through point (2, 3) and having a slope of 1 is

     (y – 3) = 1(x – 2)

=> x – y + 1 = 0

=> y – x = 1

Therefore, equation of the altitude from vertex A = y – x = 1.

Length of AD = Length of the perpendicular from A (2, 3) to BC.

The equation of BC is

      (y + 1) = {(2 + 1)/(1 - 4)}(x - 4)

=> y + 1 = -1(x - 4)

=> y + 1 = -x + 4

=> x + y – 3 = 0    ……….1

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

d = |Ax1 + By1 + C|/√(A2 + B2)

On comparing equation 1 to the general equation of line Ax + By + C = 0, we obtain

A = 1, B = 1, and C = –3

So, Length of AD = |1 * 2 + 1 * 3 - 3|/√(12 + 12) = |2|/√2 = √2 units

Thus, the equation and the length of the altitude from vertex A are y – x = 1 and √2 units

respectively.

Question 18:

If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that: 1/p2 = 1/a2 + 1/b2

Answer:

It is known that the equation of a line whose intercepts on the axes are a and b is

      x/a + y/b = 1

=> bx + ay = ab

=> bx + ay – ab = 0  ………..1

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

d = |Ax1 + By1 + C|/√(A2 + B2)

On comparing equation 1 to the general equation of line Ax + By + C = 0, we get

A = b, B = a, and C = –ab

Therefore, if p is the length of the perpendicular from point (x1, y1) = (0, 0) to line 1, we get

p = |A * 0 + B * 0 - ab|/√(b2 + a2) = |-ab|/√(b2 + a2)

On squaring both sides, we obtain

      p2 = (-ab)2/(b2 + a2)

=> p2 = a2 b2/(b2 + a2)

=> p2(b2 + a2) = a2 b2

=> (b2 + a2)/a2 b2 = 1/p2

=> 1/p2 = 1/a2 + 1/b2

Hence, we showed that 1/p2 = 1/a2 + 1/b2

 

                                                      Miscellaneous Exercise on Chapter 10

Question 1:

Find the values of k for which the line is (k - 3)x – (4 – k2)y + k2 – 7k + 6 = 0

(a) Parallel to the x-axis,      (b) Parallel to the y-axis,        (c) Passing through the origin.

Answer:

The given equation of line is

(k - 3)x – (4 – k2)y + k2 – 7k + 6 = 0   ………….1

(a) If the given line is parallel to the x-axis, then

Slope of the given line = Slope of the x-axis

The given line can be written as

(4 – k2)y = (k – 3)x + k2 – 7k + 6 = 0, which is of the form y = mx + c.

Slope of the given line = (k - 3)/(4 – k2)

Slope of the x-axis = 0

So, (k - 3)/ (4 – k2) = 0

=> k – 3 = 0

=> k = 3

Thus, if the given line is parallel to the x-axis, then the value of k is 3.

(b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be undefined.

The slope of the given line is (k - 3)/(4 – k2)

Now, (k - 3)/(4 – k2) is undefined at k2 = 4

=> k2 = 4

=> k = ±2

Thus, if the given line is parallel to the y-axis, then the value of k is ±2.

(c) If the given line is passing through the origin, then point (0, 0) satisfies the given equation

of line.

     (k - 3) * 0 – (4 – k2) * 0 + k2 – 7k + 6 = 0

=> k2 – 7k + 6 = 0   

=> (k - 6)(k - 1) = 0

=> k = 6, 1

Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.

Question 2:

Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line  √3x + y + 2 = 0.

Answer:

The equation of the given line is √3x + y + 2 = 0

This equation can be reduced as

      √3x + y + 2 = 0

=> -√3x - y = 2

On dividing both sides by √{(-√3)2 + (-1)2} = 2, we obtain

=> -√3x/2 – y/2 = 2/2

=> (-√3/2)x + (-1/2)y = 1   …………..1

Comparing equation 1 to x cos θ + y sin θ = p, we get

cos θ = -√3/2, sin θ = -1/2 and p = 1

Since the values of sin θ and cos θ are negative,

So, θ = π + π/6 = 7π/6

Thus, the respective values of θ and p are 7π/6 and 1.

 

Question 3:

Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and –6, respectively.

Answer:

Let the intercepts cut by the given lines on the axes be a and b.

It is given that

a + b = 1    ...........1

ab = –6      ...........2

On solving equations 1 and 2, we get

a = 3 and b = –2 or a = –2 and b = 3

It is known that the equation of the line whose intercepts on the axes are a and b is

       x/a + y/b = 1

=> bx + ay – ab = 0

Case I: a = 3 and b = –2

In this case, the equation of the line is –2x + 3y + 6 = 0, i.e., 2x – 3y = 6

Case II: a = –2 and b = 3

In this case, the equation of the line is 3x – 2y + 6 = 0, i.e., –3x + 2y = 6

Thus, the required equation of the lines are 2x – 3y = 6 and –3x + 2y = 6

Question 4:

What are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units.

Answer:

Let (0, b) be the point on the y-axis whose distance from line x/3 + y/4 = 1 is 4 units.

The given line can be written as

4x + 3y – 12 = 0     .............1

On comparing equation (1) to the general equation of line Ax + By + C = 0, we get

A = 4, B = 3, and C = –12

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is

given by d = |Ax1 + By1 + C|/√(A2 + B2)

Therefore, if (0, b) is the point on the y-axis whose distance from line x/3 + y/4 = 1 is 4

units, then:

      4 = |4 * 0 + 3 * b - 12|/√(42 + 32)

=> 4 = |3b - 12|/5

=> 20 = ±(3b – 120)

=> 20 = 3b – 12 or 20 = -(3b – 12)

=> 3b = 32 or 3b = -20 + 12

=> b = 32/3 or b = -8/3

Thus, the required points are (0, 32/3) or (0, -8/3).

Question 5:

Find the perpendicular distance from the origin to the line joining the points (cos θ, sin θ) and (cos ф, sin ф)

Answer:

The equation of the line joining the points (cos θ, sin θ) and (cos ф, sin ф)  is given by

       y – sin θ = {(sin ф - sin θ)/(cos ф - cos θ)}(x – sin θ)

=> (y – sin θ) (cos ф - cos θ) = (sin ф - sin θ)(x – cos θ)

=> (cos ф - cos θ)y – sin θ(cos ф - cos θ) = x(sin ф - sin θ) – cos θ(sin ф - sin θ)

=> x(sin θ - sin ф) + y(cos ф - cos θ) + cos θ sin ф – cos θ sin θ –sin θ cos ф + sin θ cos θ = 0 

=> x(sin θ - sin ф) + y(cos ф - cos θ) + sin (θ – ф) = 0

=> Ax + By + C = 0

Where A = sin θ - sin ф, B = cos ф - cos θ and C = sin (θ – ф)

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is

given by d = |Ax1 + By1 + C|/√(A2 + B2)

Therefore, the perpendicular distance (d) of the given line from point (x1, y1) = (0, 0) is

      d = |(sin θ - sin ф) * 0 + (cos ф - cos θ) * 0 + sin (θ – ф)|

                          √{(sin θ - sin ф)2 + (cos ф - cos θ)2}

=> d = |sin (θ – ф)|/ √{sin2 θ + sin2 ф - 2 sin θ * sin ф + cos2 ф + cos2 θ - 2 cos ф * cos θ}

=> d = |sin (θ – ф)|/ √{sin2 θ + cos2 θ - 2 sin θ * sin ф + sin2 ф + cos2 ф - 2 cos ф * cos θ}

=> d = |sin (θ – ф)|/ √(1 - 2 sin θ * sin ф + 1 - 2 cos ф * cos θ)

=> d = |sin (θ – ф)|/ √(2 - 2 sin θ * sin ф - 2 cos ф * cos θ)   

=> d = |sin (θ – ф)|/ √{2(1 - cos (ф - θ)}

=> d = |sin (θ – ф)|/ √{2(2sin2 (ф - θ)/2}

=> d = |sin (θ – ф)|/ |2sin (ф - θ)/2|

Question 6:

Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

Answer:

The equation of any line parallel to the y-axis is of the form

x = a     ................1

The two given lines are

x – 7y + 5 = 0 ..............2

3x + y = 0 ..............3

On solving equations (2) and (3), we get

X = -5/22 and y = 15/22

Therefore, (-5/22, 15/22) is the point of intersection of lines 2 and 3.

Since line x = a passes through point (-5/22, 15/22).

So, a = -5/22

Thus, the required equation of the line is x = -5/22

Question 7:

Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point, where it meets the y-axis.

Answer:

The equation of the given line is x/4 + y/6 = 1

This equation can also be written as

      3x + 2y – 12 = 0

=> 2y = -3x + 12

=> y = -3x/2 + 6, which is of the form y = mx + c

Now, slope of the given line = -3/2

Slope of line perpendicular to the given line = -1/(-3/2) = 2/3

Let the given line intersect the y-axis at (0, y).

On substituting x with 0 in the equation of the given line, we get

     y/6 = 1

=> y = 6

The given line intersects the y-axis at (0, 6).

The equation of the line that has a slope of 2/3 and passes through point (0, 6) is

      (y - 6) = (2/3)(x - 0)

=> 3(y - 6) = 2x

=> 3y – 18 = 2x

=> 2x – 3y + 18 = 0

Thus, the required equation of the line is 2x – 3y + 18 = 0

Question 8:

Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0.

Answer:

The equations of the given lines are

y – x = 0    ..............1

x + y = 0    ..............2

x – k = 0    ..............3

The point of intersection of lines 1 and 2 is given by

x = 0 and y = 0

The point of intersection of lines 2 and 3 is given by

x = k and y = –k

The point of intersection of lines 1 and 3 is given by

x = k and y = k

Thus, the vertices of the triangle formed by the three given lines are (0, 0), (k, –k), and (k, k).

We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is

(1/2)|x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)|

Therefore, area of the triangle formed by the three given lines

= (1/2)|0(-k – k) + k(k – 0) + k(0 + k)|

= (1/2)|k2 + k2|

= (1/2)|2k2|

= k2 square units

Question 9:

Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.

Answer:

The equations of the given lines are

3x + y – 2 = 0 ...............1

px + 2y – 3 = 0 .............2

2x – y – 3 = 0 ...............3

On solving equations 1 and 3, we obtain

x = 1 and y = –1

Since these three lines may intersect at one point, the point of intersection of lines 1 and 3 will

also satisfy line 2.

     p (1) + 2 (–1) – 3 = 0

=> p – 2 – 3 = 0

=> p = 5

Thus, the required value of p is 5.

Question 10:

If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent,

then show that m1(c2 - c3) + m2(c3 - c1) + m3(c1 - c2) = 0.

Answer:

Given equations are:

      y = m1x + c1

=> m1 x - y + c1 = 0..........1

     y = m2 x + c2

=> m2 x - y + c2 = 0..........2

    y = m3 x + c3

=> m3 x - y + c3 = 0..........3

Now three lines are concurrent if

|m1   -1    c1|

|m2   -1   c2|       = 0

|m3   -1   c3|

=> m1(-c3 + c2) - m2(-c3 + c1) + m3(-c2 + c1) = 0

=> m1(c2 - c3) + m2(c3 - c1) + m3(c1 - c2) = 0

Question 11:

Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x –2y = 3.

Answer:

Given point is (3, 2).

Equation of line is :  x - 2y = 3

 => x - 3 = 2y

=> y = x/2 – 3/2

Let the slope of required line is m.

Again given that slope between required line and x - 2y = 3

tan 45° = |(1/2 - m)/(1 + m/2)|

=> 1 =|(1/2 - m)/(1 + m/2)|

Now,

      1 = (1/2 - m)/(1 + m/2)  and -1= (1/2 - m)/(1 + m/2)    (since |x| = -x and x)

=> 1 + m/2 = 1/2 - m and -(1 + m/2) = (1/2 - m)

=> m + m/2 = 1/2 -1 and -1 - m/2 = 1/2 - m

=> 3m/2 = -1/2 and m - m/2 = 1+ 1/2

=> m = -1/3 and m/2 = 3/2

=> m =-1/3 and m = 3

Case 1: m = -1/3 and point is (3, 2)

      y - 2 = (-1/3)(x - 3)

=> 3(y - 2) = -x + 3

=> 3y - 6 = -x + 3

=> 3y + x = 3 + 6

=> x + 3y = 9

Case 2: m = 3 and point is (3, 2)

      y - 2 = 3(x - 3)

=> y-2 = 3x - 9

=> 3x - y = 9-2

=> 3x - y = 7

Thus, the equations of the lines are 3x – y = 7 and x + 3y = 9.

Question 12:

Find the equation of the line passing through the point of intersection of the lines

4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.

Answer:

Given equation of lines are :

      4x + 7y - 3 = 0 .............1

and 2x - 3y + 1 = 0 .............2

Let the required equation of lines having equal intercept is

     x/a + y/a  = 1

=> x + y = a ..........3

After solving equation 1 and 2, we get

x = 1/13, = 5/13

So, (1/3, 5/13) is the point of intersection of the given lines.

Since the equation 3 passes through this point, So

     1/13 + 5/13 = a

=> a = 6/13

Now put this value in equation 3, we get

      x + y = 6/13

=> 13(x + y) = 6

=> 13x + 13y = 6

So, the equation of line is 13x + 13y = 6

Question 13:

Show that the equation of the line passing through the origin and making an angle θ

with the line y = mx + c is y/x = (m ± tan θ)/(1 ∓ m * tan θ)

Answer:

Let the equation of the line passing through the origin be y = m1x.

If this line makes an angle of θ with line y = mx + c, then angle θ is given by

So, tan θ = |(m1 – m)/(1 + m1m)|

=> tan θ = |(y/x – m)/(1 + ym/x)|

=> tan θ = ±{(y/x – m)/(1 + ym/x)}

=> tan θ = (y/x – m)/(1 + ym/x) or tan θ = -(y/x – m)/(1 + ym/x)

Case 1:

      tan θ = (y/x – m)/(1 + ym/x)

=> tan θ(1 + ym/x) = (y/x – m)

=> tan θ + (ym/x) tan θ = y/x – m

=> m + tan θ = (y/x)(1 – m * tan θ)

=> y/x = (m + tan θ)/ (1 – m * tan θ)

Case 2:

      tan θ = -(y/x – m)/(1 + ym/x)

=> tan θ(1 + ym/x) = -(y/x – m)

=> tan θ + (ym/x) tan θ = -y/x + m

=> m + tan θ = (y/x)(1 + m * tan θ)

=> y/x = (m + tan θ)/ (1 + m * tan θ)

Therefore, the required line is given by y/x = (m ± tan θ)/(1 ∓ m * tan θ)

Question 14:

In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?

Answer:

Given line is x + y = 4 .............1

and points are: (-1, 1) and (5, 7)

Equation of line which joins the points

       y - 1 = {(7 - 1)/(5 + 1)}*(x + 1)

=> y-1 = (6/6)*(x + 1)

=> y - 1 = x + 1

=> x - y + 1 + 1 = 0

=> x – y + 2 = 0 ..........2

After solving equation 1 and 2, we get

x = 1, y = 3

So, (1, 3) is the point of intersection of these two lines.

Now let point (1, 3) divide the line joining (-1, 1) and (5, 7) in the ratio 1 : k

According to section formula,

      (1, 3) = [{k(-1) + 1 * 5}/(1 + k), (k * 1 + 1 * 7)/(1 + k)]

=> (1, 3) = {(-k + 5)/(1 + k), (k + 7)/(1 + k)}

Now,

        (-k + 5)/(1 + k) = 1

  => -k + 5 = 1 + k

  => 5 - 1= k + k

  => 4 = 2k

  => k = 2

Hence, the required ratio is 1 : 2

Question 15:

Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.

Answer:

The equation of line AB is 4x + 7y + 5 = 0   …….1

and equation of line PQ is 2x - y = 0   ………2

Class_11_Straight_Lines_LineSegment3

Solve the equations 1 and 2 for obtaining point Q.

From equation 2 put y = 2x in equation 1

4x + 7 * 2x + 5 = 0

=> 4x + 14x + 5 = 0

=> 18x + 5 = 0

=> x = -5/18

Put it in equation 2, we get

y = 2x = -5/9

So, the co-ordinate of Q = (-5/18, -5/9)

Now, Length of PQ = √{(x2 - x1)² + (y2 - y1)²}

                                   = √{(1 + 5/18)² + (2 + 5/9)²}

                                   = √{(23/18)² + (23/9)²}

                                   = √{23²{1/18² + 1/9²}

                                   = 23√{1/9²(1/4 + 1)}

                                   =23/9 √(1/4 + 1)

                                   = 23/9 √(5/4)

                                   = 23√5/18 unit

Question 16:

Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection

with the line x + y = 4 may be at a distance of 3 units from this point.

Answer:

Let the straight line is y = mx + c

This line passes through the point (-1, 2)

So, 2 = -m + c

=> c = m + 2

Now y = mx + m + 2 .........1

Given equation of line is : x + y = 4 ..........2

Solving equation 1 and 2, we get

x = (2 - m)/(m + 1) and y = (5m + 1)/(m + 1)

Now this point is at a distance of 3 units from the point (-1, 2)

From distance formula

√[{(2 - m)/(m + 1) + 2}2 + {(5m + 2)/(m + 1) - 2}2]  = 3

Squaring both side, we get

      {(2 - m)/(m + 1) + 1}2 + {(5m + 2)/(m + 1) - 2}2 = 9

=> {(2 – m + m + 1)/(m + 1)}2 + {(5m + 2 - 2m - 2)/(m + 1)}2 = 9

=> 9/(m + 1)2 + 9m2/(m + 1)2 = 0

=> (1 + m2)/(m + 1)2 = 1                   (divide by 9 on both side)

=> 1 + m2 = (m + 1)2

=> 1 + m2 = (m2 + 2m +1)

=> 1 + m2 = m2 + 2m + 1

=> 2m = 0

=> m = 0

Since the slope of the line is zero.

Hence, the line is parallel to x-axis.

Question 18:

Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

Answer:

Given equation of line is  

    x + 3y = 7 .........1

Class_11_Straight_Lines_LineSegment5

Let the point (a, b) is the image of point (3, 8)

Slope of line AB = (b - 8)/(a - 3)

Since the given line x + 3y = 7 is perpendicular to AB

So (b - 8)/(a - 3) *(-1/3)  = -1    (Since product of two slopes = -1)

=> (b - 8)/(a - 3) = 3

=> (b - 8) = 3(a - 3)

=> b - 8 = 3a - 9

=> 3a - b = 9 - 8

=> 3a - b = 1 .........2

Now mid-point of AB = {(a + 3)/3, (b + 8)/2}

This point also satisfies equation 1

So, (a + 3)/2 + 3(b + 8)/2 = 7

=> a + 3 + 3b + 24 = 2*7

=> a + 3b + 27 = 14

=> a + 3b = 14-27

=> a + 3b = -13 ......3

After solving equation 2 and 3, we get

a = -1, b = -4

Hence, the image of the point (3, 8) with respect to x + 3y = 8 is (-1, -4).

Question 19:

If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

Answer:

Given equation of line are:

y = 3x + 1 ..............1

Slope = 3

    2y = x + 3

=> y = x/3 + 3/2 ...........2

Slope = 1/3

y = mx + 4 ........3

Slope = m

Given that equation 1 and 2 are equally inclined with equation 3.

So, the angle between equation 1 and 3 = angle between equation 2 and 3

=> |(3 - m)/(1 + 3m)| = |(1/2 - m)/(1 + m/2)|

=> |(3 - m)/(1 + 3m)| = |(1 - 2m)/(2 + m)|

=> (3 - m)/(1 + 3m) = (1 - 2m)/(2 + m) and (3 - m)/(1 + 3m) = -(1 - 2m)/(2 + m)

=> (3 - m)(2 + m) = (1 - 2m)(1 + 3m) and (3 - m)*(2 + m) = -(1 + 3m)(1 - 2m)

=> 6 + 3m - 2m - m2 = 1 + 3m - 2m - 6 m2 and 6 + 3m - 2m - m2 = -(1 + 3m - 2m - 6 m2)

=> -m2 + m + 6 = 1 + m - 6m2 and -m2 + m + 6 = -1 – m + 6m2

=> 5m2 + 5 = 0 and 7m2 - 2m - 7 = 0

=> m2 + 1= 0

=> m2 = -1 which is not possible.

Again, 7m2 - 2m - 7 = 0

=> m = {2 + √(4 - 4*7*(-7))}/(2*7) and m = {2 - √(4 - 4*7*(-7))}/(2*7)

=> m = {2 + √(4 + 4*49)}/14 and m = {2 - √(4 + 4*49)}/14

=> m = {2 + 2√(1 + 49)}/14 and m = {2 - 2√(1 + 49)}/14

=> m = {1 + √50}/7 and m = {1 - √50}/7

=> m = {1 + 5√2}/7 and m = {1 - 5√2}/7

Hence, the value of m is {1 ± 5√2}/7

Question 20:

If sum of the perpendicular distances of a variable point P (x, y) from the lines

x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.

Answer:

Given Point is P(x, y).

Lines are x + y - 5 = 0 and 3x - 2y + 7 =0

Now distance of P(x, y) from the line x + y - 5 = 0 is

d1 = (x + y - 5)/√(1 + 1) = (x + y - 5)/√2

Again distance of P(x, y) from the line 3x - 2y + 7 = 0 is

d2 = (3x - 2y + 7)/√{32 + (-2)2} = (3x - 2y + 7)/√(9 + 4) = (3x - 2y + 7)/√13

Given sum of distances is equal to 10

=> d1 + d2 = 10

=> (x + y - 5)/√2 + (3x - 2y + 7)/√13 = 10

=> √13*(x + y - 5) + √2*(3x - 2y + 7) = 10*√2*√13

=> x(√13 + 3√2) + y(√13 - 2√2) + (7√2 - 5√13) = 10

=> y(√13 - 2√2) = -(√13 + 3√2)x + 10 - (7√2 - 5√13)

=> y =  {-(√13 + 3√2)/(√13 - 2√2)}x + {10 - (7√2 - 5√13)}/(√13 - 2√2)

This forms a line y = mx + c

Hence, the given point P(x, y) must move on a line.

Question 21:

Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and  3x + 2y + 6 = 0.

Answer:

Given equation are:

9x + 6y - 7 = 0 .........1

3x + 2y + 6 = 0.........2

Now equation of line which is equidistant from both the lines is

      |(9x + 6y - 7)/√(92 + 62 )| = |(3x + 2y + 6)/√(32 + 22)|

=> |(9x + 6y - 7)/√(81+36 )| = |(3x + 2y + 6)/√(9 + 4 )|

=> |(9x + 6y - 7)/√117| = |(3x + 2y + 6)/√13|

=> |(9x + 6y - 7)/√117| = |(3x + 2y + 6)/√13|

=> |(9x + 6y - 7)/√(13*9)| = |(3x + 2y + 6)/√13|

=> |(9x + 6y - 7)/(3*√13)| = |(3x + 2y + 6)/√13|

=> |(9x + 6y - 7)/3| = |3x + 2y + 6|

=> (9x + 6y - 7)/3 = 3x + 2y + 6  and  (9x + 6y - 7)/3 = -(3x + 2y + 6)

=> 9x + 6y - 7 = 3*(3x + 2y + 6) and 9x + 6y - 7 = -3*(3x + 2y + 6)

=> 9x + 6y - 7 = 9x + 6y + 18 and 9x + 6y - 7 = -9x - 6y - 18

Now, 9x + 6y - 7 = 9x + 6y + 18 which is not possible.

Again, 9x + 6y - 7 = -9x - 6y - 18

=> 9x + 6y - 7 + 9x + 6y + 18 = 0

=> 18x + 12y + 11 = 0

Hence, the required equation is: 18x + 12y + 11 = 0

Question 22:

A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the

reflected ray passes through the point (5, 3). Find the coordinates of A.

Answer:

Let the incident ray strike x-axis at the point A and the coordinate is (x, 0).

Now, from the figure

      tan θ = (3 - 0)/(5 - x)

 => tan θ = 3/(5 - x) .............1

Again slope of the incident ray 

     tan(π - θ) = (0 - 2)/(x - 1)

=> - tan θ = -2/(x - 1)

=> tan θ = 2/(x - 1) .........2

From equation 1 and 2,

      3/(5 - x) = 2/(x - 1)

=> 3(x - 1) = 2(5 - x)

=> 3x - 3 = 10 - 2x

=> 3x + 2x = 10 + 3

=> 5x = 13

=> x = 13/5

Hence, the required point is (13/5, 0)

Question 23:

Prove that the product of the lengths of the perpendiculars drawn from the points                             

{√(a2 - b2), 0} and {-√(a2 - b2), 0} to the line x cos θ /a + y sin θ /b = 1 is b2.

Answer:

The equation of the given line is

      x cos θ /a + y sin θ /b = 1

=> bx cos θ + ay sin θ – ab = 0    …………….1

Length of the perpendicular from the point {√(a2 - b2), 0} to the line 1 is

p1 = |b cos θ * √(a2 - b2) + a sin θ * 0 - ab|/√(b2 cos2 θ + a2 sin2 θ)

      = |b cos θ * √(a2 - b2) - ab|/√(b2 cos2 θ + a2 sin2 θ)   …….2

Length of the perpendicular from the point {-√(a2 - b2), 0} to the line 1 is

p2 = |b cos θ * {-√(a2 - b2)} + a sin θ * 0 - ab|/√(b2 cos2 θ + a2 sin2 θ)

      = |b cos θ * √(a2 - b2) + ab|/√(b2 cos2 θ + a2 sin2 θ)  ……….3

Multiply equation 2 and 3, we get

p1p2 = {|b cos θ * √(a2 - b2) - ab|* |b cos θ * √(a2 - b2) + ab|}/{√(b2 cos2 θ + a2 sin2 θ)}2

         = {|b cos θ * √(a2 - b2) - ab|* |b cos θ * √(a2 - b2) + ab|}/(b2 cos2 θ + a2 sin2 θ)

         = {|b cos θ * √(a2 - b2)} 2 – (ab)2|/(b2 cos2 θ + a2 sin2 θ)    

         = |b2 cos2 θ * (a2 - b2) – a2 b2|/(b2 cos2 θ + a2 sin2 θ) 

         = |a2 b2 cos2 θ – b4 cos2 θ – a2 b2|/(b2 cos2 θ + a2 sin2 θ)

         = b2|a2 cos2 θ – b2 cos2 θ – a2|/(b2 cos2 θ + a2 sin2 θ)  

         = b2|a2 cos2 θ – b2 cos2 θ – a2 sin2 θ - a2 cos2 θ|/(b2 cos2 θ + a2 sin2 θ)

         = b2|-(b2 cos2 θ + a2 sin2 θ)|/(b2 cos2 θ + a2 sin2 θ)

         = b2(b2 cos2 θ + a2 sin2 θ)/(b2 cos2 θ + a2 sin2 θ)

         = b2

Hence, p1p2 = b2 

Question 24:

A person standing at the junction (crossing) of two straight paths represented by the equations

2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time.

Find equation of the path that he should follow.

Answer:

Given equations are:

2x - 3y + 4 = 0 .............1

3x + 4y - 5 = 0..............2

After solving it, we get

x = -1/17, y = 22/17

Suppose that person is standing at point P(-1/17, 22/17)

Again given line is 6x - 7y + 8 = 0

Slope = 6/7

Slope of perpendicular line = -7/6

Now equation of line passing through (-1/17, 22/17) and perpendicular to 6x - 7y + 8 = 0 is

      y - 22/17 = (-7/6)(x + 1/17)

=> 6(17y - 22)/17 = (-7)*(17x + 1)/17

=> 6(17y - 22) = (-7)*(17x + 1)

=> 102y - 132 = -119x - 7

=> 119x + 102y - 132 + 7 = 0

=> 119x + 102y - 125 = 0

This is the required equation of line.

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