Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B The Straight Line Solutions Exercise 3(b) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B The Straight Line Solutions Exercise 3(b)

I.

Question 1.

Find the sum of the squares of the inter¬cepts of the line 4x – 3y = 12 on the axes, of co-ordinates.

Solution:

Equation of the given line is

\(\frac{4 x}{12}-\frac{3 y}{12}=1\)

\(\frac{x}{3}+\frac{y}{-4}=1\)

a = 3, b = -4

Sum of the squares = a² + b²

= 9 + 16 = 25

Question 2.

If the portion of a straight line inter¬cepted between the axes of co-ordinates is bisected at (2p, 2q), write the equation of the straight line.

Solution:

Equation of AB in the intercept form is

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

Co-ordinates of A are (a, 0) and B are (0, b)

M is the mid-point of AB

Co-ordinates of M are (\(\frac{a}{2}\), \(\frac{b}{2}\)) = (2p, 2q)

\(\frac{a}{2}\) = 2p, \(\frac{b}{2}\) = 2q

a = 4p, b = 4q

Substituting in (1), equation of AB is

\(\frac{x}{4 P}+\frac{y}{4 Q}=1\)

\(\frac{x}{P}+\frac{y}{Q}=4\)

Question 3.

If the linear equation ax + by + c = 0

(a,b,c ≠ 0) and lx + my + n = 0

represent the same line and r = \(\frac{l}{a}\) = \(\frac{n}{c}\)

write the value of r in terms of m and b.

Solution:

ax + by + c — 0 and

lx + my + n = 0 represent the same line

∴ \(\frac{1}{a}=\frac{m}{b}=\frac{n}{c}=r\)

\(\frac{m}{b}\) = r

Question 4.

Find the angle made by the straight line y = – √3x + 3 with the positive direction of the X-axis mea-sured in the counter clock-wise direction.

Solution:

Equation of the given line isy = -√3x + 3 Suppose a is the angle made by this line with positive X – axis in the counter clock – wise direction.

tan α = – √3 = tan \(\frac{2 \pi}{3}\)

α = \(\frac{2 \pi}{3}\)

Question 5.

The intercepts of a straight line on the axes of co-ordinates are a and b. If p is the length of the perpendicular drawn from the origin to this line. Write the value of p in terms of a and b.

Solution:

Equation of the line in the intercept form is

p = length of the perpendicular from origin

II.

Question 1.

In what follows, p denotes the distance of the straight line from the origin and a denotes the angle made by the normal ray drawn from the origin to the straight line with \(\stackrel{\leftrightarrow}{O X}\) measured in the anticlockwise sense. Find the equations of the straight lines with the following values of p and a.

i) p = 5, a = 60°

ii) p = 6, a = 150°

iii) p = 1, α = \(\frac{7 \pi}{4}\)

iv) p = 4, α = 90°

v) p = 0, α = 0

vi) p = 2√2, α = \(\frac{5 \pi}{4}\)

Solution:

Equation of the line in the normal form is x cos α + y sin α = p

i) p = 5, α = 60°

cos α = cos 60° = \(\frac{1}{2}\)

sin α = sin 60° = \(\frac{\sqrt{3}}{2}\)

Equation of the line is x. \(\frac{1}{2}\) + y. \(\frac{\sqrt{3}}{2}\) = 5

⇒ x + √3y= 10

ii) p = 6, α = 150°

cos α = cos 150° = cos (180° – 30°)

= -cos 30° = – \(\frac{\sqrt{3}}{2}\)

sin α = sin 150°

= sin (180° – 30°)

= sin 30° = \(\frac{1}{2}\)

Equation of the line is

x.(-\(\frac{\sqrt{3}}{2}\)) + y.\(\frac{1}{2}\) = 6

-√3x + y = 12

or √3x – y + 12 = 0

iii) p = 1, α = \(\frac{7 \pi}{4}\)

cos α = cos 315° = cos (360° – 45°)

= cos 45° = \(\frac{1}{\sqrt{2}}\)

sin α = sin 315° = sin (360° – 45°)

= -sin45° = \(\frac{1}{\sqrt{2}}\)

Equation of the line is

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

x – y = √2

x – y – √2 = 0

iv) p = 4, α = 90°

cos α = cos 90° = 0, sin α = sin 90° = 1

Equation of the line is

x.0 + y.1 = 4

y = 4

v) p = 0, α = 0

cos α = cos 0 = 1, sin α = sin 0 = 0

Equation of the line is

x.1 + y.0 = 0

x = 0

vi) p = 2√2 , α = \(\frac{5 \pi}{4}\)

cos α = cos 225° = cos (180° + 45°)

= -cos 45° = – \(\frac{1}{\sqrt{2}}\)

sin α = sin 225° = sin (180° + 45°)

=-sin 45° = – \(\frac{1}{\sqrt{2}}\)

Equation of the line is

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

– x – y = 4

or x + y + 4 = 0

Question 2.

Find the equations of the straight line in the symmetric form, given the slope and a point on the line in each part of the question.

i) √3, (2, 3)

ii) –\(\frac{1}{\sqrt{3}}\), (-2, 0)

iii) -1, (1,1)

Solution:

i) Equation of the line in the symmetric form is

\(\frac{x-x_{1}}{\cos \alpha}=\frac{y-y_{1}}{\sin \alpha}=r\)

(x_{1}, y_{1}) = (2, 3)

m = tan α = √3 ⇒ α = 60°

cos α = cos 60° = \(\frac{1}{2}\)

sin α = sm 60° = \(\frac{\sqrt{3}}{2}\)

Equation of the line in symmetric form is

\(\frac{x-2}{\cos \frac{\pi}{3}}=\frac{y-3}{\sin \frac{\pi}{3}}\)

ii) (x_{1}, y_{1}) = (-2, 0)

tan α = –\(\frac{1}{\sqrt{3}}\) ⇒ α = 180° – 30° = 150°

Equation of the line is \(\frac{x+2}{\cos 150^{\circ}}=\frac{y}{\sin 150^{\circ}}\)

iii) tan α = -1, α =180°- 45° = 135°

(x_{1}, y_{1}) = (1, 1)

Equation of the line is \(\frac{x-1}{\cos \left(\frac{3 \pi}{4}\right)}=\frac{y-1}{\sin \left(\frac{3 \pi}{4}\right)}\)

Question 3.

Transform the following equation into

a) Slope-intercept form

b) Intercept form and

c) Normal form

i) 3x + 4y = 5

ii) 4x – 3y + 12 = 0

iii) √3x + y = 4

iv) x + y + 2 = 0

v) x + y – 2 = 0

vi) √3x + y + 10 = 0

Solution:

i) 3x + 4y = 5

Slope-intercept form

4y = -3x + 5

\(y=\left(-\frac{3}{4}\right) x+\left(\frac{5}{4}\right)\)

Intercept form :

3x + 4y = 5

x cos α + y sin α = 1

ii) 4x – 3y + 12 = 0

Slope-intercept form :

3y = 4x + 12

y = (\(\frac{4}{3}\))x + 4

Intercept form:

4x – 3y+ 12 = 0

-4x + 3y = 12

\(\frac{-4 x}{12}+\frac{3 y}{12}=1\)

\(\frac{x}{(-3)}+\frac{y}{4}=1\)

Normal form :

4x – 3y + 12 = 0

– 4x + 3y = 12

x cos α + y sin α = \(\frac{12}{5}\)

iii) √3x + y = 4

Slope-intercept form :

√3x + y = 4

y = -√3x + 4

Intercept form :

√3x + y = 4

iv) x + y + 2 = 0

Slope-intercept form

x + y+.2 = 0

y = -x – 2

= (-1)x + (-2)

Intercept form:

x + y + 2 = 0

-x – y = 2

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

Normal form:

x + y + 2 = 0

-x – y = 2

v) x + y – 2 = 0

Slope-intercept form :

x + y – 2 = 0

y = – x + 2

Intercept form:

x + y – 2 = 0

x + y = 2

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

Normal form :

x + y-2 = 0

x + y = 2

vi) √3x + y + 10 = 0

Slope-intercept form :

√3x + y + 10 = 0

y = -√3x – 10

Intercept form :

√3x + y = -10

Normal form :

√3x + y = -10

Dividing with \(\sqrt{3+1}=2\), we get

\(\frac{-\sqrt{3}}{2} \cdot x+\frac{-1}{2} \cdot y=5\)

x cos 30° + y sin 30° = 5

Question 4.

If the product of the intercepts made by the straight line x tan α + y sec α = 1 (0 ≤ α < \(\frac{\pi}{2}\)), on the co-ordinate axes is equal to sin α, find α.

Solution:

Equation of the line is x tan α + y sec α = 1

\(\frac{x}{\cot \alpha}+\frac{y}{\cos \alpha}=1\)

a = cot α, b = cos α

Given ab = sin α

cot α. cos α = sin α

\(\frac{\cos ^{2} \alpha}{\sin \alpha}\) = sin α ⇒ cos² α = sin α

tan² α = 1 ⇒ tan α = ±1

α = 45°

Question 5.

If the sum of the reciprocals of the intercepts made by a variable straight line on the axes of co-ordinate is a constant, then prove that the line always passes through a fixed point.

Solution:

Equation of the line in the intercept form is

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

Sum of the reciprocals of the intercepts

The line (1) passes through the fixed point

\(\left(\frac{1}{k}, \frac{1}{k}, \frac{1}{k}\right)\)

Question 6.

Line L has intercepts a and b on the axes of coordinates. When the axes are rota¬ted through a given angle, keeping the origin fixed, the same line L has intercepts p and q on the transformed axes. Prove that \(\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{P^{2}}+\frac{1}{Q^{2}}\).

Solution:

Equation of the line in the old system in the intercept form is

\(\frac{x}{a}+\frac{y}{b}=1 \Rightarrow \frac{x}{a}+\frac{y}{b}-1=0\)

Length of the perpendicular from origin

Equation of the line in the second system in the intercept form is

\(\frac{x}{p}+\frac{y}{Q}=1 \Rightarrow \frac{x}{p}+\frac{y}{Q}-1=0\)

Length of the perpendicular from origin

Since the origin and the given line remain unchanged we have from (1) and (2)

Question 7.

Transform the equation \(\frac{x}{a}+\frac{y}{b}=1\) into the normal form when a > 0 and b > 0. If the perpendicular distance of the straight line from the origin is p, deduce that \(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\).

Solution:

III.

Question 1.

A straight line passing through A(-2, 1) makes an angle 30° with \(\overrightarrow{O X}\) in the positive direction. Find the points on the straight line whose distance from A is 4 units.

Solution:

Co-ordinates of any point on the given line are (x_{1} + r cos α, y_{1} + r sin α)

α = 30° ⇒ cos α = cos 30° = \(\frac{\sqrt{3}}{2}\),

sin α = sin = 30° = \(\frac{1}{2}\)

(x_{1}, y_{1}) = (-2, 1) ⇒ x_{1} = -2 y_{1} = 1

Taking r = 4 ⇒ co-ordinates of P are

Question 2.

Find the points on the line 3x – 4y – 1 = 0 which are at a distance of 5 units from the point (3, 2).

Solution:

Equation of the line in the symmetric form is

\(\frac{x-3}{\cos \alpha}=\frac{y-2}{\sin \alpha}=r\)

Co-ordinates of the point P are

(3 + r cos α, 2 + r sin α) = (3 + 5 cos α, 2 + 5 sin α)

P is a point on 3x – 4y – 1 = 0

3(3 + 5 cos α) – 4(2 + 5 sin α) -1 = 0

9 + 15 cos α – 8 – 20 sin α – 1 = 0

15 cos α – 20 sin α = 0

15 cos α = + 20 sin α

tan α = + T

Case i) : cos α = +\(\frac{4}{5}\), sin α \(\frac{3}{5}\)

Case ii) : cos α = –\(\frac{4}{5}\), sin α \(\frac{3}{5}\)

Case i) : Co-ordinates of P are

\(\left(3+5 \cdot \frac{4}{5}, 2+5 \cdot \frac{3}{5}\right)=(7,5)\)

Case ii) : Co-ordinates of P are

\(\left(3-5 \cdot \frac{4}{5}, 2-5 \cdot \frac{3}{5}\right)=(-1,-1)\)

Question 3.

A straight line whose inclination with the positive direction of the X-axis measured in the antidock wise sense is it/3 makes positive intercept on the Y-axis. If the straight line is at a distance of 4 from the origin, find its equation.

Solution:

Given α = π/3, p = 4

m = tan α = tan 60° = √3

Equation of the line in the slope – intercept form is

y = √3 x + c

√3x – y + c = 0

Distance from the origin = 4

\(\frac{|0-0+c|}{\sqrt{3+1}}=4\)

|c| = 4 × 2 = 8

c = ± 8

Given c > 0 ∴ c = 8

Equation of the line is √3x – y + 8 = 0

Question 4.

A straight line L is drawn through the point A (2, 1) such that its point of intersection with the straight line x + y = 9 is at a distance of 3√2 from A. Find the angle which the line L makes With the positive direction of the X – axis.

Solution:

Suppose a is the angle made by L with the positive X – axis

Any point on the line is

(x_{1} + r cos α_{1}, y_{1} + r sin α) = (2 + 3√2 cos α 1 + 3√2 sin α)

This is a point on the line x + y = 9

2+ 3√2 cos α+ 1 + 3√2 sin α = 9

3√2 (cos α + sin α) = 6

cos α + sm α = \(\frac{6}{3 \sqrt{2}}\) = √2

\(\frac{1}{\sqrt{2}}\). cos α + \(\frac{1}{\sqrt{2}}\) sin α = 1

cos α. cos 45° + sin α. sin 45° = 1

cos (α – 45°) = cos 0°

α – 45° = 0 ⇒ α = 45° = \(\frac{\pi}{3}\)

Question 5.

A straight line L with negative slope passes through the point (8, 2) and cuts positive co-ordinate axes at the points P and Q. Find the minimum value of OP + OQ as L varies, when O is the origin.

Solution:

Equation of the line passing through A(8, 2) with negative slope ‘-m’ is y – 2 = -m(x – 8)

mx + y – (2 + 8m) = 0

mx + y = 2 + 8m

OP =X Intercept = \(\frac{2+8 m}{m}\)

OQ = Y – Intercept = 2 + 8m 2 + 8m

For f(m) to have minimum or maximum, we must have f'(m) = 0

and ‘f’ has minimum at m = \(\frac{1}{2}\)

∴ Minimum value of f(m) = Minimum Value I of OP + OQ at m = \(\frac{1}{2}\)

∴ Minimum value of OP + OQ as L varies, where ‘O’ is the origin is 18.