Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 10 Straight Lines Ex 10.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 10 Straight Lines Ex 10.3

Question 1.

Reduce the following equations into slope-intercept form and find their slopes and the intercept:

(i) x + 7y = 0

(ii) 6x + Sy – 5 = 0

(iii) y = 0

Solution:

(i) x + 7y = 0 or y = – \(\frac{1}{7}\)x + 0.

âˆ´ Slope = – \(\frac{1}{7}\) and y-intercept = 0.

(ii) 6x + 3y – 5 = 0 or 3y = – 6x + 5

y = – 2x + \(\frac{5}{3}\).

âˆ´ Slope = – 2 and y – intercept = \(\frac{5}{3}\).

(iii) y = 0 or y = 0x + 0.

âˆ´ Slope = 0, y-intercept = 0.

Question 2.

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

(i) 3x + 2y – 12 = 0

(ii) 4x – 3y = 6

(iii) 3y + 2 = 0

Solution:

(i) 3x + 2y – 12 = 0 or 3x + 2y = 12.

Dividing by 12, we get

\(\frac{x}{4}\) + \(\frac{y}{6}\) = 1.

âˆ´ Intercepts on the axes are 4 and 6.

(ii) 4x – 3y = 6 or \(\frac{4}{6}\)x – \(\frac{y}{2}\) = 1.

or \(\frac{x}{3/2}\) + \(\frac{y}{-2}\) = 1.

Intercepts on the axes are \(\frac{3}{2}\) and – 2.

(iii) 3y + 2 = 0 or 3y = – 2.

or \(\frac{y}{-2/3}\) = 1 or \(\frac{x}{0}\) + \(\frac{y}{-2/3}\) = 1.

Intercepts on the axes are 0 and – \(\frac{2}{3}\).

Question 3.

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

(i) x – \(\sqrt{3}\)y + 8 = 0

(ii) y – 2 = 0

(iii) x – y = 4

Solution:

(i) x – \(\sqrt{3}\)y + 8 = 0 or x – \(\sqrt{3}\)y = – 8.

or – x + \(\sqrt{3}\)y = 8.

Put rcos Ï‰ = – 1 and rsin Ï‰ = \(\sqrt{3}\).

Squaring and adding r^{2} = 1 + 3 = 4, âˆ´ r = 2.

cos Ï‰ = \(\frac{1}{2}\), sin Ï‰ = \(\frac{\sqrt{3}}{2}\) âˆ´ Ï‰ = 120Â° = \(\frac{2Ï€}{3}\).

âˆ´ xcos Ï‰ + y sin Ï‰ = \(\frac{8}{2}\) = 4.

âˆ´ p = 4 and Ï‰ = \(\frac{2Ï€}{3}\).

(ii) y – 2 = 0

Comparing it with Ax + By = C, we get

A = 0, B = 1, r = \(\sqrt{0+1^{2}}\) = 1.

âˆ´ cos Ï‰ = 0, sin Ï‰ = \(\frac{8}{2}\) = 4.

âˆ´ y = 2 â‡’ x cos \(\frac{Ï€}{2}\) + ysin \(\frac{Ï€}{2}\) = 2

âˆ´ Ï‰ = \(\frac{Ï€}{2}\), p = 2.

(iii) x – y = 4 ………………… (1)

Put r cos Ï‰ = 1, r sin Ï‰ = – 1.

So, r = \(\sqrt{1^{2}+(-1)^{2}}\) = \(\sqrt{2}\)

and cos Ï‰ = \(\frac{1}{\sqrt{2}}\), sin Ï‰ = \(\frac{1}{\sqrt{2}}\).

â‡’ Ï‰ = 360Â° – 45Â° = 315Â°.

Dividing (1) by \(\sqrt{2}\), we get

\(\frac{1}{\sqrt{2}}\)x + (- \(\frac{1}{\sqrt{2}}\)y) = \(\frac{y}{\sqrt{2}}\) = 2\(\sqrt{2}\).

âˆ´ Normal form of the given line is

xcos 315Â° + ysin 315Â° = 2\(\sqrt{2}\).

âˆ´ Ï‰ = 315Â°, p = 2\(\sqrt{2}\).

Question 4.

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

Solution:

The given line is

12(x + 6) = 5(y – 2)

or 12x + 72 = 5y – 10

or 12x – 5y + 72 + 10 = 0

or 12x – 5y + 82 = 0.

âˆ´ The perpendicular distance from (x_{1}, y_{1}) to the line ax + by + c = 0 is \(\frac{\left(a x_{1}+b y_{1}+c\right)}{\sqrt{a^{2}+b^{2}}}\).

The point (x_{1}, y_{1}) is (- 1, 1).

âˆ´ Perpendicular distances from (- 1, 1) to the line 12x – 5y + 82 = 0

Question 5.

Find the points on the x-axis, whose distances from \(\frac{x}{3}\) + \(\frac{y}{4}\) = 1 are 4 units.

Solution:

The given line is \(\frac{x}{3}\) + \(\frac{y}{4}\) = 1.

Multiplying by 12, we get

4x + 3y = 12 or 4x + 3y – 12 = 0.

Any point on the x-axis is (x, 0).

âˆ´ Perpendicular distance from (x, 0) to the given line

= \(\frac{|4 x-12|}{\sqrt{4^{2}+3^{2}}}\) = 4.

or \(\frac{|4 x-12|}{\sqrt{25}}\) = 4 or |4x – 12| = 5 Ã— 4 = 20.

Taking (+ve) sign, 4x – 12 = 20 or 4x = 20 + 12 = 32

â‡’ x = \(\frac{32}{4}\) = 8.

Taking (- ve) sign, 4x – 12 = – 20.

â‡’ 4x = – 20 + 12 = – 8

âˆ´ x = – 2.

âˆ´ Required points on the x-axis are (8, 0) and (- 2, 0).

Question 6.

Find the distance between the parallel lines:

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

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

Solution:

One of the lines is

15x + 8y – 34 = 0.

Putting y = 0, 15x = 34 or x = \(\frac{34}{15}\).

âˆ´ The point (\(\frac{34}{15}\), 0) lies on 15x + 8y – 34 = 0.

âˆ´ Perpendicular distance from (\(\frac{34}{15}\), 0) to 15x + 8y + 31 = 0 is

Alternative method:

Perpendicular distance from the origin of these lines are p_{1} = \(\frac{-34}{\sqrt{15^{2}+8^{2}}}\) = \(\frac{- 34}{17}\) and p_{2} = \(\frac{31}{17}\).

âˆ´ Perpendicular distance between the given parallel lines = |p_{2} – p_{1}|

= |\(\frac{-34}{17}\) – \(\frac{31}{17}\)| = |\(\frac{-65}{17}\)| = \(\frac{65}{17}\).

(ii) The parallel lines are

l(x + y) + p = 0, lx + ly – r = 0.

Perpendicular distances from origin are

âˆ´ Perpendicular distance between the given parallel lines are

|\(\frac{p}{\sqrt{2l}}\) – \(\frac{- r}{\sqrt{2l}}\)| = \(\frac{p+r}{\sqrt{2l}}\).

Question 7.

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

Solution:

The given line is 3x – 4y + 2 = 0.

âˆ´ Slope of the line = \(\frac{3}{4}\).

âˆ´ Slope of a parallel line = \(\frac{3}{4}\).

âˆ´ The line passing through (- 2, 3) and parallel to 3x – 4y + 2 = 0 is

y – 3 = \(\frac{3}{4}\)(x + 2).

or 4y – 12 = 3x + 6

or 3x – 4y + 12 + 6 = 0 or 3x – 4y + 18 = 0.

Alternative Method.

Any line parallel to 3x – 4y + 2 = 0 is 3x – 4y = k.

[Note that the coeffs. of x and y are the same for parallel lines]

The line passes through (- 2, 3).

So, 3(- 2) – 4(3) = k

or k = – 6 – 12 = – 18

âˆ´ Equation of the required line is 3x – 4y = – 18

or 3x – 4y + 18 = 0.

Question 8.

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

Solution:

The line having the x-intercept 3, passes through the point (3, 0).

The given line is x – 7y + 5 = 0.

âˆ´ Slope of it = \(\frac{1}{7}\).

âˆ´ Slope of perpendicular line = – 7.

âˆ´ The line passing through (3, 0) and slope – 7 is

y – 0 = – 7(x – 3)

or y = – 7x + 21

âˆ´ 7x + y – 21.

Alternative Method.

Any line perpendicular to ax + by + c = 0 is bx – ay + k – 0.

Note that here coefficients of x and y are interchanged and sign between them also changes from + to – and – to + in between.

âˆ´ The line perpendicular to x – 7y + 5 = 0 is

7x + y = k.

Putting x = 3, y = 0 in it, we get 21 = k.

âˆ´ Equation of the required line is

7x + y = 21.

Question 9.

Find the angles between the lines \(\sqrt{3x}\) + y = 1 and z + \(\sqrt{3}\)y = 1.

Solution:

The first line is \(\sqrt{3}\)x + y = 1 or y = – \(\sqrt{3x}\) + 1.

âˆ´ The slope m_{1} = – \(\sqrt{3}\)

âˆ´ m_{1} = – \(\sqrt{3}\).

The other line is x + \(\sqrt{3y}\) = 1.

or y = – \(\frac{1}{\sqrt{3}}\)x + \(\frac{1}{\sqrt{3}}\).

âˆ´ The slope m_{2} of the line x + \(\sqrt{3y}\) = 1 is – \(\frac{1}{\sqrt{3}}\).

âˆ´ m_{2} = – \(\frac{1}{\sqrt{3}}\).

If Î¸ is the angle between two lines, then

So, acute angle between the lines is \(\frac{Ï€}{6}\).

Question 10.

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

Solution:

Slope of the line PQ passing through P(h, 3) and Q(4, 1).

= \(\frac{3-1}{h-4}\) = \(\frac{2}{h-4}\).

Slope of the line AB, i.e; 7x – 9y – 19 = 0 is \(\frac{7}{9}\).

The lines AB and PQ are perpendicular to each other.

âˆ´ m_{1}m_{2} = – 1 or \(\frac{2}{h-4}\) Ã— \(\frac{7}{9}\) = – 1.

âˆ´ 14 = – 9(h – 4) = – 9h + 36

âˆ´ 9h = 36 – 14 = 22.

âˆ´ h = \(\frac{22}{9}\).

Question 11.

Prove that the line through the point (x_{1}, y_{1}) and parallel to the line Ax + By + C = 0 is

A(x – x_{1}) + B(y – y_{1}) = 0.

Proof:

The given line is Ax + By + C = 0

of y = – \(\frac{A}{B}\)x – \(\frac{C}{B}\)

âˆ´ Slope of the parallel line = – \(\frac{A}{B}\)

The parallel line passes through (x_{1},Â y_{1}) is

y – y_{1} = – \(\frac{A}{B}\)(x – x_{1})

or B(y – y_{1}) = – A(x – x_{1})

or A(x – x_{1}) + B(y – y_{1}) = 0.

Alternatively. The line parallel to Ax + By + C = 0 is

Ax + By + k = 0 ………………. (1)

It passes through (x_{1}y_{1})

âˆ´ Ax_{1} + By_{1} + k = 0 ……………… (2)

Subtracting (2) from (1), we get

A(x – x_{1}) + B(y – y_{1}) = 0.

Question 12.

Two lines passing through the point (2, 3) intersects each other at an angle of 60Â°. If slope of one line is 2, find the equation of the other line.

Solution:

Let m be the slope of the other line.

Slope of one line = 2.

Angle between them = 60Â°.

Equation of the line passing through P(2, 3) with slope

or (1 + 2\(\sqrt{3}\))y – 3(1 + 2\(\sqrt{3}\)) = (2 – \(\sqrt{3}\)x – 2(2 – \(\sqrt{3}\)).

or (1 + 2\(\sqrt{3}\))y – 3 – 6\(\sqrt{3}\) = (2 – \(\sqrt{3}\))x – 4 + 2\(\sqrt{3}\)

âˆ´ (2 – \(\sqrt{3}\))x – (1 + 2\(\sqrt{3}\))y – 4 + 2\(\sqrt{3}\) + 3 + 6\(\sqrt{3}\) = 0.

or (2 – \(\sqrt{3}\))x – (1 + 2\(\sqrt{3}\))y – 1 + 8\(\sqrt{3}\) = 0.

âˆ´ Required lines are

(\(\sqrt{3}\) + 2)x + (2\(\sqrt{3}\) – 1)y – 8\(\sqrt{3}\) – 1 = 0.

and (2 – \(\sqrt{3}\))x – (1 + 2\(\sqrt{3}\))y + 8\(\sqrt{3}\) – 1 = 0.

Question 13.

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

Solution:

Slope of the line joining the points A(- 1, 2) and B(3, 4)

= \(\frac{4-2}{3+1}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\).

Right bisector PQ is perpendicular to AB.

âˆ´ Slope of PQ = – 2.

Middle point of AB is (\(\frac{-1+3}{2}\), \(\frac{2+4}{2}\)), i.e; (1, 3).

Right bisector passes through M(1, 3).

âˆ´ Equation of right bisector PQ is

y – 3 = – 2(x – 1) = – 2x + 2.

or 2x + y – 3 – 2 = 0

or 2x + y – 5 = 0.

Question 14.

Find the co-ordinates of the foot of the perpendicular from a point (- 1, 3) to the line 3ax – 4y – 16 = 0.

Solution:

The equation of the given line is 3x – Ay – 16 = 0.

The equation of a line perpendicular to the given line is

4x + 3y + k = 0, where k is a constant.

If this line passes through the point (- 1, 3), then

– 4 + 9 + k = 0 â‡’ k = – 5.

âˆ´ The equation of the line passing through the point (- 1, 3) and perpendicular to the given line is

4x + 3y – 5 = 0.

âˆ´ The required foot of the perpendicular is the point of intersection of the lines

3x – 4y – 16 = 0 …………… (1)

and 4x + 3y – 5 = 0 ……………….. (2)

Solving (1) and (2), by cross-multiplication, we have:

Question 15.

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

Solution:

Let the perpendicular OM is drawn from the origin to AB. M is the foot of the perpendicular.

Slope of OM = \(\frac{2-0}{-1-0}\) = \(\frac{2}{-1}\).

Slope of AB = m

OM âŠ¥ AB.

âˆ´ m Ã— (- 2) = – 1

âˆ´ m = \(\frac{1}{2}\).

M(- 1, 2) lies on y = mx + c or y = \(\frac{1}{2}\)x + c.

âˆ´ 2 = \(\frac{1}{2}\) Ã— (- 1) + c, c = 2 + \(\frac{1}{2}\) = \(\frac{5}{2}\).

âˆ´ m = \(\frac{1}{2}\) and c = \(\frac{5}{2}\).

Question 16.

If p and q are the lengths of perpendiculars from the origin to the lines xcos Î¸ – ysin Î¸ = kcos 2Î¸ and x sec Î¸ + y cosec Î¸ = k respectively, prove that p^{2} + 4q^{2} = k^{2}.

Solution:

The perpendicular distance from the origin to the line xcos Î¸ – ysin Î¸ = kcos 2Î¸ ………………. (1)

âˆ´ The perpendicular distance q from the origin to the line (2) is

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.

Solution:

The vertices of âˆ† ABC are A(2, 3), B(4, – 1) and C(1, 2) and AM is the altitude.

Slope of BC = \(\frac{2+1}{1-4}\) = \(\frac{3}{-3}\) = – 1.

âˆ´ Slope of altitude AM = 1. [âˆµ m_{1}m_{2} = – 1]

Now, altitude passes through A(2, 3) and has the slope 1.

âˆ´ Equation of AM is

y – 3 = 1(x – 2).

or x – y + 3 – 2 = 0

or y – x = 1.

Equation of BC passing through B(4, – 1) and C(1, 2) is

y + 1 = – (x – 4).

or x + y = 3

or x + y – 3 = 0.

âˆ´ Length of altitude = AM

= Perpendicular distance from A(2, 3) to BC

= \(\frac{2+3-3}{\sqrt{1^{2}+1^{2}}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)

âˆ´ Equation of altitude is y – x = 1 and length of altitude = \(\sqrt{2}\).

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 \(\frac{1}{p^{2}}\) = \(\frac{1}{a^{2}}\) + \(\frac{1}{b^{2}}\).

Solution:

Equation of the line which makes intercepts a and b on the axes is

\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1.

âˆ´ The perpendicular distance p from the origin is given by