Inter 1st Year Maths Straight Lines Solutions Exercise 9b

Practicing the AP Board Solutions Class 11 Maths and Chapter 9 Inter 1st Year Maths Straight Lines Solutions Exercise 9b Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Straight Lines Solutions Exercise 9b

Straight Lines Exercise 9b Solutions

Straight Lines Class 11 Exercise 9b Solutions – Straight Lines 9b Exercise Solutions

I. In exercises 1 to 7, find the equation of the line that satisfies the given conditions:

Question 1.
Write the equations for the X-axis, Y-axis.
Solution:
The y-coordinate of every point on the x-axis is 0.
∴ The equation of the x-axis is y = 0
The x-coordinate of the y-axis is y = 0
∴ The equation of the y-axis is y = 0

Question 2.
Passing through the point (-4, 3) with slope \(\frac {1}{2}\).
Solution:
Given, Point (-4, 3) and slope, m = \(\frac {1}{2}\)
We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀) only if its coordinates satisfy the equation y – y₀ = m(x – x₀)
So, y – 3 = \(\frac {1}{2}\)(x – (-4))
⇒ y – 3 = \(\frac {1}{2}\)(x + 4)
⇒ 2(y – 3) = x + 4
⇒ x + 4 – (2y – 6) = 0
⇒ x + 4 – 2y + 6 = 0
⇒ x – 2y + 10 = 0
∴ The equation of the line is x – 2y + 10 = 0.

Question 3.
Passing through (0, 0) with slope m.
Solution:
Given Point (0, 0) and slope m = m
We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀) only if its coordinates satisfy the equation y – y₀ = m(x – x₀)
So, y – 0 = m(x – 0)
⇒ y = mx
⇒ y – mx = 0
∴ The equation of the line is y – mx = 0

Question 4.
Passing through (2, 2√3) and inclined with the x-axis at an angle of 75°.
Solution:
Given point (2, 2√3) and θ = 75°
Equation of line (y – y₁) = m(x – x₁)
where, m = slope of line = tan θ
and (x₁, y₁) are the points through which the line passes
∴ m = tan 75°
75° = 45° + 30°
Applying the formula:

We know that the point (x, y) lies on the line with slope m through the fixed point (x₁, y₁), only if its coordinates satisfy the equation y – y₁ = m(x – x₁)
Then, y – 2√3 = (2 + √3) (x – 2)
⇒ y – 2√3 = 2x – 4 + √3x – 2√3
⇒ y = 2x – 4 + √3x
⇒ (2 + √3)x – y – 4 = 0
∴ The equation of the line is (2 + √3)x – y – 4 = 0

Question 5.
Intersecting the x-axis at a distance of 3 units to the left of the origin with a slope of -2.
Solution:
Given Slope, m = -2
We know that if a line L with slope m makes x-intercept d, then the equation of L is y = m(x – d)
If the distance is 3 units to the left of the origin, then d = 3
So, y = (-2) (x – (-3))
⇒ y = (-2) (x + 3)
⇒ y = -2x – 6
⇒ 2x + y + 6 = 0
∴ The equation of the line is 2x + y + 6 = 0

Question 6.
Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with the positive direction of the x-axis.
Solution:
Given θ = 30°
We know that slope, m = tan θ
m = tan 30° = \(\frac{1}{\sqrt{3}}\)
We know that the point (x, y) on the line with slope m and y-intercept c lies on the line only if y = mx + c
If the distance is 2 units above the origin, c = +2
So, y = \(\frac{1}{\sqrt{3}}\)x + 2
⇒ y = \(\frac{x+2 \sqrt{3}}{\sqrt{3}}\)
⇒ √3y = x + 2√3
⇒ x – √3y + 2√3 = 0
∴ The equation of the line is x – √3y + 2√3 = 0.

Question 7.
Passing through the points (-1, 1) and (2, -4).
Solution:
Given Points (-1, 1) and (2, -4)
We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by
y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
⇒ y – y₁ = \(\frac{-4-1}{2-(-1)}\) (x – (-1))
⇒ y – y₁ = \(\frac {-5}{3}\) (x + 1)
⇒ 3(y – 1) = (-5) (x + 1)
⇒ 3y – 3 + 5x + 5 = 0
⇒ 5x + 3y + 2 = 0
∴ The equation of the line is 5x + 3y + 2 = 0.

Question 8.
The vertices of ΔPQR are P(2, 1), Q(-2, 3) and R(4, 5). Find the equation of the median through the vertex R.
Solution:
Given, Vertices of ΔPQR,
i.e., P(2, 1), Q(-2, 3), and R(4, 5)
Let RL be the median of vertex R
So, L is a midpoint of PQ
We know that the midpoint formula is given by \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
∴ L = \(\left(\frac{2-2}{2}, \frac{1+3}{2}\right)\) = (0, 2)
We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by
y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
⇒ y – 5 = \(\frac{2-5}{0-4}\) (x – 4)
⇒ y – 5 = \(\frac {-3}{-4}\) (x – 4)
⇒ (-4) (y – 5) = (-3) (x – 4)
⇒ -4y + 20 = -3x + 12
⇒ -4y + 20 + 3x – 12 = 0
⇒ 3x – 4y + 8 = 0
∴ The equation of the median through the vertex R is 3x – 4y + 8 = 0.

Question 9.
Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6).
Solution:
Given Points are (2, 5) and (-3, 6).
We know that slope,

We know that two non-vertical lines are perpendicular to each other only if their slopes are negative reciprocals of each other.
Then, m = \(\frac {-1}{m}\) = \(\frac{\frac{-1}{-1}}{5}\) = 5
We know that the point (x, y) lies on the line with slope m through the fixed point (x0, y0), only if its coordinates satisfy the equation y – y₀ = m(x – x₀)
Then, y – 5 = 5(x – (-3))
⇒ y – 5 = 5x + 15
⇒ 5x + 15 – y + 5 = 0
⇒ 5x – y + 20 = 0
∴ The equation of the line is 5x – y + 20 = 0.

Question 10.
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.
Solution:
We know that the coordinates of a point dividing the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio m : n are

We know that slope,

We know that two non-vertical lines are perpendicular to each other only if their slopes are negative reciprocals of each other.
Then, m = (-1/m) = -1/3
We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), only if its coordinates satisfy the equation y – y₀ = m(x – x₀)
Here, the point is \(\left(\frac{2+n}{1+n}, \frac{3}{1+n}\right)\)
⇒ \(\left(y-\frac{3}{1+n}\right)=\frac{-1}{3}\left(x-\frac{2+n}{1+n}\right)\)
⇒ 3((1 + n)y – 3) = (-(1 + n)x + 2 + n)
⇒ 3(1 + n)y – 9 = -(1 + n)x + 2 + n
⇒ (1 + n)x + 3(1 + n)y – n – 9 – 2 = 0
⇒ (1 + n)x + 3(1 + n)y – n – 11 = 0
∴ The equation of the line is (1 + n)x + 3(1 + n)y – n – 11 = 0

Question 11.
Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).
Solution:
Given that the line cuts off equal intercepts on the coordinate axes,
i.e., a = b
We know that the equation of the line intercepts a and b on the x-axis and the y-axis, respectively, which is
\(\frac{x}{a}+\frac{y}{b}\) = 1
So, \(\frac{x}{a}+\frac{y}{a}\) = 1
x + y = a …………(1)
Given point (2, 3)
2 + 3 = a
a = 5
Substituting the value of ‘a’ in (1), we get
x + y = 5
x + y – 5 = 0
∴ The equation of the line is x + y – 5 = 0.

Question 12.
If the portion of a straight line intercepted between the axes of co-ordinates is bisected at (2p, 2q), write the equation of the straight line.
Solution:

Let a, b be the intercepts of the line
The line cuts x-axis at A(a, 0), y-axis at B(0, b)
Midpoint of \(\overline{\mathrm{AB}}\) is \(\left(\frac{a}{2}, \frac{b}{2}\right)\)
\(\left(\frac{a}{2}, \frac{b}{2}\right)\) = (2p, 2q)
⇒ \(\frac {a}{2}\) = 2p; \(\frac {b}{2}\) = 2q
⇒ a = 4p, b = 4q
∴ Equation of the line is \(\frac{x}{4 p}+\frac{y}{4 q}\) = 1
⇒ \(\frac{x}{p}+\frac{y}{q}\) = 4

Question 13.
Find the angle made by the straight line y = -√3x + 3 with the positive direction of the X-axis measured in the counterclockwise direction.
Solution:
Let θ be the inclination, slope = -√3
⇒ tan θ = -√3
⇒ θ = \(\frac{2 \pi}{3}\)

Question 14.
The intercepts of a straight line on the axes of coordinates 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 having intercepts a, b is \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ bx + ay = ab
⇒ \(\frac{b}{\sqrt{a^2+b^2}} x+\frac{a}{\sqrt{a^2+b^2}} y=\frac{a b}{\sqrt{a^2+b^2}}\)
P = length of the perpendicular from (0, 0) to \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ P = \(\frac{1-11}{\sqrt{\left(\frac{1}{a^2}\right)+\left(\frac{1}{b^2}\right)}}=\frac{|a b|}{\sqrt{a^2+b^2}}\)

II.

Question 1.
Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Solution:
We know that the equation of the line making intercepts a and b on the x-axis and the y-axis, respectively, is
\(\frac{x}{a}+\frac{y}{b}\) = 1 ……….(1)
Given: sum o’f intercepts = 9
a + b = 9
b = 9 – a
Now, substitute the value of b in the above equation, and we get
\(\frac{x}{a}+\frac{y}{9-a}\) = 1
Given the line passes through point (2, 2)

18 = a(9 – a)
18 = 9a – a²
a² – 9a + 18 = 0
Upon factorising, we get
a² – 3a – 6a + 18 = 0
a(a – 3) – 6(a – 3) = 0
(a – 3) (a – 6) = 0
a = 3 (or) a = 6
Let us substitute in (1)
Case (i): a = 3
Then b = 9 – 3 = 6
\(\frac{x}{3}+\frac{y}{6}\) = 1
2x + y = 6
2x + y = 6
2x + y – 6 = 0
Case(ii): a = 6
theh b = 9 – 6 = 3
\(\frac{x}{6}+\frac{y}{3}\) = 1
x + 2y = 6
x + 2y – 6 = 0
∴ The equation of the line 2x + y – 6 = 0 (or) x + 2y – 6 = 0

Question 2.
Find the equation of the line through the point (0, 2) making an angle \(\frac{2 \pi}{3}\) with the positive x-axis. Also, find the equation of a line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Solution:
Given Point (0, 2) and θ = \(\frac{2 \pi}{3}\)
we know that m = tan θ
m = tan(\(\frac{2 \pi}{3}\)) = √3
We know that the point (x, y) lies on the line with slope m through the fixed point (x0, y0) only if its coordinates satisfy the equation y – y₀ = m(x – x₀)
y – 2 = -√3(x – 0)
y – 2 = -√3x
√3x + y – 2 = 0
Given, the equation of the line parallel to the above obtained equation crosses the y-axis at a distance of 2 units below the origin.
So, the point = (0, -2) and m = -√3
From point-slope form equation,
y – (-2) = -√3(x – 0)
y + 2 = -√3x
√3x + y + 2 = 0
∴ The equation of the line √3x + y – 2 = 0, and the line parallel to it is √3x + y + 2 = 0.

Question 3.
The perpendicular from the origin to a line meets it at the point (-2, 9). Find the equation of the line.
Solution:
Given Points are origin (0, 0) and (-2, a)
We know that slope

We know that two non-vertical lines are perpendicular to each other only if their slopes are negative reciprocals of each other.
m = \(\frac{-1}{m}=\frac{-1}{\frac{-9}{2}}=\frac{2}{9}\)
We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀) only if its coordinates satisfy the equation y – y₀ = m(x – x₀)
⇒ 9(y – 9) = 2(x + 2)
⇒ 9y – 81 = 2x + 4
⇒ 2x + 4 – 9y + 81 = 0
⇒ 2x – 9y + 85 = 0
∴ The equation of the line is 2x – 9y + 85 = 0

Question 4.
The length L (in centimetres) 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.
Solution:
Let us assume ‘L’ along the x-axis and ‘c’ along the y-axis;
We have two points (124.942, 20) and (125.134, 110) in the xy-plane.
We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

Question 5.
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?
Solution:
Assuming the relationship between the setting price and demand is linear.
Let us assume the selling price per litre along the x-axis and demand along the y-axis.
We have two points (14, 980) and (16, 1220) in the xy-plane.
We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

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
∴ The owner can sell 1340 litres weekly at Rs. 17/litres.

Question 6.
P(a, b) is the midpoint of a line segment between axes. Show that the equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 2.
Solution:
Let AB be a line segment whose midpoint is P(a, b).
Let the coordinates of A and B be (0, y) and (x, 0), respectively.

We know that the midpoint is given by \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
Since P is the midpoint of (a, b)
\(\left(\frac{0+x}{2}, \frac{y+0}{2}\right)\) = (a, b)
\(\left(\frac{x}{2}, \frac{y}{2}\right)\) = (a, b)
a = \(\frac {x}{2}\) and b = \(\frac {y}{2}\)
x = 2a and y = 2b
A = (0, 2b) and B = (2a, 0)
We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

Hence proved.

Question 7.
Point R(h, k) divides a line segment between the axes in the ratio 1 : 2. Find the equation of the line.
Solution:
Let us consider AB to be the line segment, such that R(h, k) divides it in the ratio 1 : 2.
So, the coordinates of A and B are (0, y), and (x, 0), respectively.

We know that the coordinates of a point dividing the line segment joining the points (x₁, y₁) and (x₂, y₂) internally in the ratio m : n are

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

3h(y – 3k) = -6kx
3hy – 9hk = -6kx
6kx + 3hy = 9hk
Let us divide both sides by 9hk, and we get,
\(\frac{2 x}{3 h}+\frac{y}{3 k}\) = 1
∴ The equation of the line is given by \(\frac{2 x}{3 h}+\frac{y}{3 k}\) = 1

Question 8.
By using the concept of the equation of a line, prove that the following three points are collinear.
(i) (3, 0), (-2, -2), and (8, 2)
(ii) (-5, 1), (5, 5), (10, 7)
(iii) (a, b + c), (b, c + a), (c, a + b)
Solution:
(i) According to the question,
If we have to prove that the given three points (3, 0), (-2, -2), and (8, 2) are collinear,
then we have to also prove that the line passing through the points (3, 0) and (-2, -2) also passes through the point (8, 2).
By using the formula, the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

-5y = -2(x – 3)
-5y = -2x + 6
2x – 5y = 6
If 2x – 5y = 6 passes through (8, 2).
2x – 5y = 2(8) – 5(2)
= 16 -10
= 6
= RHS
The line passing through points (3, 0) and (-2, -2) also passes through the point (8, 2).
Hence proved the given three points are collinear.

(ii) Let A(-5, 1), B(5, 5), C(10, 7)
Slope of \(\overline{\mathrm{AB}}=\frac{5-1}{5+5}=\frac{4}{10}=\frac{2}{5}\)
Equation of the line \(\overline{\mathrm{AB}}\) is (y – 1) = \(\frac {2}{5}\)(x + 5)
⇒ 5y – 5 = 2x + 10
⇒ 2x – 5y + 15 = 0
2(10) – 5(7) + 15 = 0
⇒ C lies on \(\overline{\mathrm{AB}}\).
∴ A, B, and C are collinear.

(iii) Let A(a, b + c), B(b, c + a), C(c, a + b)
Slope of \(\overline{\mathrm{AB}}=\frac{\mathrm{c}+\mathrm{a}-\mathrm{b}-\mathrm{c}}{\mathrm{~b}-\mathrm{a}}=\frac{\mathrm{a}-\mathrm{b}}{-(\mathrm{a}-\mathrm{b})}\) = -1
Equation of the line \(\overline{\mathrm{AB}}\) is y – (b + c) = -1(x – a)
⇒ x + y = a + b + c
c + (a + b) = a + b + c
⇒ c lies on \(\overline{\mathrm{AB}}\).
∴ A, B, and C are collinear.