OP Malhotra Class 10 Maths Solutions Chapter 11 Coordinate Geometry Ex 11(b)

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S Chand Class 10 ICSE Maths Solutions Chapter 11 Coordinate Geometry Ex 11(b)

Question 1.
Find the slope of a line whose inclination to the positive direction of x-axis in anticlockwise direction is given as :
(a) 30°
(b) 45°
(c) 60°
(d) 15°
(e) 75°
Solution:
If the inclination of a line to the positive direction is θ then slope of that line will be tan θ. Therefore
(a) tan 30° = \(\frac{1}{\sqrt{3}}\)
(b) tan 45° = 1
(c) tan 60° = \(\sqrt{3}\)
(d) tan 15° = 0.2679
(From the table of tan θ)
(e) tan 75° = 3.7231
(From the table of tan θ)

Question 2.
Find the slope and inclination of the line through each pair of the following points:
(a) (1, 2) and (5, 6)
(b) (0, 0) and (\(\sqrt{3}\), 3)
Solution:
(a) Points are (1, 2) and (5, 6)
∴ Slope (m) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{5-1}=\frac{4}{4}\) = 1
Angle of inclination = 45° (∵ tan 45° = 1)

(b) Points are (0, 0) and (\(\sqrt{3}\), 3)
∴ Slope (m) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{3-0}{\sqrt{3}-0}=\frac{3}{\sqrt{3}}\)
= \(\frac{3 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{3 \sqrt{3}}{3}=\sqrt{3} .\)
Angle of inclination = 60° (∵ tan 60° = \(\sqrt{3}\))

Question 3.
The side BC of an equilateral ∆ABC is parallel to the x-axis. What are the slopes of its sides?
Solution:
In an equilateral ∆ABC, each angle = 60°
and side BC || x-axis

Sides AB and AC are produced to meet the x-axis is at L and M
∴ ∠L = ∠B = 60°
(∵ Each angle of the triangle is 60°)
and ∠AMX = 120°
∴ Slope of AB = tan 60° = \(\sqrt{3}\)
and slope of AC = tan 120° = tan (180° – 60°) = tan 60° = – \(\sqrt{3}\)
and slope of BC = tan 0° = 0 (∵ BC || x-axis)

Question 4.
In a regular hexagon ABCDEF, AB || ED || the x-axis. What are the slopes of its sides?
Solution:
ABCDEF is a regular hexagon in which side AB and ED are parallel to x-axis.
Now slope of AB = tan 0° = 0

Slope of BC = tan 60° = \(\sqrt{3}\)
Slope of CD = tan 120° = – \(\sqrt{3}\)
Slope of DE = tan 0° = 0 (∵ ED || x-axis)
Slope of EF = tan 60° = \(\sqrt{3}\)
and slope of FA = tan 120° = – \(\sqrt{3}\)

Question 5.
Find y if the slope of the line joining (- 8, 11), (2, y) is \(\frac { -4 }{ 3 }\).
Solution:
We know that slope (m) = \(\frac{y_2-y_1}{x_2-x_1}\)
⇒ \(\frac{-4}{3}=\frac{y-11}{2-(-8)} \Rightarrow \frac{-4}{3}=\frac{y-11}{2+8}\)
⇒ \(\frac{-4}{3}=\frac{y-11}{10}\) ⇒ 3y – 33 = – 40
⇒ 3y = – 40 + 33 ⇒ 3y = – 7
∴ y = \(\frac { -7 }{ 3 }\)

Question 6.
Find the value of a, if the line passing through (-5, -8) and (3, 0) is parallel to the line passing through (6, 3) and (4, a).
Solution:
Slope (m₁) of the line passing through the points (-5, -8) and (3, 0)
= \(\frac{y_2-y_1}{x_2-x_1}=\frac{0-(-8)}{3-(-5)}\)
= \(\frac{8}{3+5}=\frac{8}{8}\) = 1
and slope (m₂) of the line passing through the points (6, 3) and (4, a) = \(\frac{a-3}{4-6}=\frac{a-3}{-2}\)
∵ The two lines are parallel
∴ Their slopes are equal
⇒ m₁ = m₂
⇒ 1 = \(\frac { a – 3 }{ – 2 }\) ⇒ – 2 = a – 3
⇒ a = – 2 + 3 = 1
Hence a = 1

Question 7.
Find the slope of a line perpendicular to the line whose slope is :
(a) 3
(b) 1
(c) 5
(d) – 5
(e) 0
(f) Infinite
Solution:
We know that the slope of a line is m, then slope of the line perpendicular to the given line will be \(\frac { -1 }{ m }\)
(a) Slope of the line (m) = \(\frac { 1 }{ 3 }\)
∴ Slope of its perpendicular = \(\frac { -1 }{ m }\) = \(\frac { -3 }{ 1 }\) = – 3

(b) Slope of the line (m) = \(\frac { -5 }{ 6 }\)
∴ Slope of its perpendicular = \(\frac { -1 }{ m }\) = – (\(\frac { -6 }{ 5 }\)) = \(\frac { 6 }{ 5 }\)

(c) Slope of the line (m) = 5
∴ Slope of its perpendicular = \(\frac { -1 }{ m }\) = \(\frac { -1 }{ 5 }\)

(d) Slope of the line (m) = -5\(\frac { 1 }{ 7 }\) = \(\frac { -36 }{ 7 }\)
∴ Slope of its perpendicular = \(\frac { -1 }{ m }\) = – (\(\frac { -7 }{ 36 }\)) = \(\frac { 7 }{ 36 }\)

(c) Slope of the line (m) = 0
∴ Slope of its perpendicular = \(\frac { -1 }{ m }\) = \(\frac { -1 }{ 0 }\) = infinite

(d) Slope of the line (m) = infinite
∴ Slope of its perpendicular = \(\frac { -1 }{ m }\) = \(\frac { -1 }{ infinite }\) = 0

Question 8.
Find the slope of a line perpendicular to the line which passes through the pair of the following points :
(a) (0, 8) and (-5, 2);
(b) (1, -11) and (5, 2)
(c) (-k, h) and (b, -f)
(d) (x₁, y₁) and (x₂, y₂)
Solution:
(a) Points are (0, 8) and (-5, 2)
∴ Slope of the line joining these points (m)
= \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{2-8}{-5-0}=\frac{-6}{-5}=\frac{6}{5}\)
∴ Slope of the line perpendicular to it = \(\frac { -1 }{ m }\) = \(\frac { -5 }{ 6 }\)

(b) Points are (1, -11) and (5, 2)
Slope of the line joining these points
= m = \(\frac{y_2-y_1}{x_2-x_1}=\frac{2-(-11)}{5-1}=\frac{2+11}{5-1}=\frac{13}{4}\)
∴ Slope of the line perpendicular to it = \(\frac { -1 }{ m }\) = \(\frac { -4 }{ 13 }\)

(c) Points are (-k, h) and (b, -j)
Slope of the line joining these points
= m = \(\frac{y_2-y_1}{x_2-x_1}=\frac{-f-h}{b-(-k)}=\frac{-(f+h)}{b+k}\)
∴ Slope of the line perpendicular to it = \(\frac { -1 }{ m }\) = \(\frac{-(b+k)}{-(f+h)}=\frac{b+k}{f+h}\)

(d) Points are (x₁, y₁) and (x₂, y₂)
Slope of the line joining these points = m
= \(\frac{y_2-y_1}{x_2-x_1}\)
∴ Slope of the line perpendicular to it = \(\frac { -1 }{ m }\) = – \(\frac{x_2-x_1}{y_2-y_1}\)

Question 9.
In rectangle ABCD, the slope of AB = \(\frac { 5 }{ 6 }\). State the slope of
(a) BC
(b) CD
(c) DA
Solution:
In rectangle ABCD,
Slope of AB = \(\frac { 5 }{ 6 }\), Therefore

Question 10.
In parallelogram ABCD, slope of AB = – 2, slope of BC = \(\frac { 3 }{ 5 }\). State the slope of
(a) AD
(b) CD
(c) the altitude of AD
(d) the altitude of CD
Solution:
In parallelogram ABCD, AB || CD and DA || BC
Slope of AB = – 2 and slope of BC = \(\frac { 3 }{ 5 }\)
(a) ∵ DA || BC
∴ Slope of AD = Slope of BC = \(\frac { 3 }{ 5 }\)

(b) ∵ CD || AB
∴ Slope of CD = Slope of AB = – 2

(c) Slope of altitude of AD = – \(\left(\frac{1}{\text { Slope of } A D}\right)\)

(d) Slope of the altitude of CD = \(\frac{-1}{\text { Slope of } C D}\)
= – (\(\frac { 1 }{ -2 }\)) = \(\frac { 1 }{ 2 }\)

Question 11.
The vertices of a AABC are A (1, 1), B (7, 3) and C (3, 6). State the slope of the altitude to
(a) AB
(b) BC
(c) AC.
Solution:
Vertices of a ∆ABC are A (1, 1), B (7, 3) and C (3, 6)

(a) ∴ Slope of AB = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{3-1}{7-1}=\frac{2}{6}=\frac{1}{3}\)
and slope of its altitude = \(\frac{-1}{\text { Slope of } \mathrm{AB}}\)
= \(\frac { -3 }{ 1 }\) = – 3
Similarly,

(b) Slope of BC = \(\frac{6-3}{3-7}=\frac{3}{-4}\)
and slope of its altitude BC = \(\frac{-1}{\text { Slope of } \mathrm{BC}}\)
= – \(\left(\frac{-4}{3}\right)=\frac{4}{3}\)

(c) Slope of AC = \(\frac{6-1}{3-1}=\frac{5}{2}\)
∴ slope of its altitude AC = \(\frac{-1}{\text { Slope of } \mathrm{AC}}\)
= \(\frac { -2 }{ 5 }\)

Question 12.
The line joining (-5, 7) and (0, -2) is perpendicular to the line joining (1, -3) and (4, x). Find x.
Solution:
Slope of line joining the points (-5, 7) and (0, – 2) = m₁ = \(\frac{y_2-y_1}{x_2-x_1}=\frac{-2-7}{0-(-5)}=\frac{-9}{5}\)
and slope of the line joining the points
(1, – 3) and (4, x) = m₂ = \(\frac{x-(-3)}{4-1}=\frac{x+3}{3}\)
∵ These lines are perpendicular to each other
∴ m₁ x m₂ = – 1
⇒ \(\frac{-9}{5} \times \frac{x+3}{3}\) = 1
⇒ \(\frac{-3(x+3)}{5}=-1 \Rightarrow x+3=-1 \times \frac{5}{-3}=\frac{5}{3}\)
⇒ x = \(\frac{5}{3}-3=\frac{5-9}{3}=\frac{-4}{3}\)
Hence x = = \(\frac { -4 }{ 3 }\)

Question 13.
The vertices of a quad. PMQS are P (0, 0), M (3, 2), Q (7, 7) and S (4, 5). Show that PMQS is a parallelogram.
Solution:
Vertices of a quad PMQS are P(0,0), M (3, 2) , Q (7, 7) and S (4, 5)
Slope of PM (m₁) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{2-0}{3-0}=\frac{2}{3}\)
Similarly,
Slope of MQ = (m₂) = \(\frac{7-2}{7-3}=\frac{5}{4}\)
Slope of QS (m₃) = \(\frac{5-7}{4-7}=\frac{-2}{-3}=\frac{2}{3}\)
and slope of SP = \(\frac{5-0}{4-0}=\frac{5}{4}\)
∵ Slope of PM = Slope of QS = \(\frac { 2 }{ 3 }\)
∴ PM || QS … (i)
Similarly slope of MQ = slope of SP = \(\frac { 5 }{ 4 }\)
∴ MQ || SP … (ii)
∴ From (i) and (ii)
PMQS is a parallelogram

Question 14.
Show that P (a, b), Q (a + 3, b + 4),R (a – 1, b + 7), S (a – 4, b + 3) are the vertices of a square. What is the area of the square?
Solution:
Co-ordinates of P = (a, b), Q (a – 3, b + 4). R(a – 1, b + 1), S(a – 4, b + 3)

∵ All the sides PQ, QR, RS and SP are equal and both diagonals PR and QS are also equal
∴ PQRS is a square
Now area of square PQRS = (side)² = (PR)² = (5)² = 25 square units.

Question 15.
Without using Pythagoras theorem, show that the points A (0, 4), B (1, 2) and C (3, 3) are the vertices of a right angled triangle.
Solution:
Vertices of a AABC are A (0, 4), B (1, 2) and C (3, 3)
Now slope of AB (m₁) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{2-4}{1-0}\)
= \(\frac { -2 }{ 1 }\) = – 2
Similarly,
Slope of BC (m₂) = \(\frac{3-2}{3-1}=\frac{1}{2}\)
and slope of CA (m₃) = \(\frac{4-3}{0-3}=\frac{1}{-3}\)
∵ m₁ x m₂ = – 2 x \(\frac { 1 }{ 2 }\) = – 1
∴ AB and BC are perpendicular to each other
∴ ∆ABC is a right angled triangle.

Question 16.
If the points (x, 1), (1, 2) and (0, y + 1) are collinear, show that \(\frac { 1 }{ x }\) + \(\frac { 1 }{ y }\) = 1.
Solution:
Let points are A (x, 1), B (1, 2) and C (0, y + 1)
∴ Slope of AB = \(\frac{y_2-y_1}{x_2-x_1}=\frac{2-1}{1-x}=\frac{1}{1-x}\)
∵ Similarly, slope of BC = \(\frac{y+1-2}{0-1}=\frac{y-1}{-1}\)
∴ A, B and C are collinear
∴ Slope of AB = slope of BC
⇒ \(\frac{1}{1-x}=\frac{y-1}{-1}\) ⇒ (1 – x) (y – 1) = – 1
⇒ y – 1 – xy + x = – 1
⇒ x + y = -1 + 1 + xy ⇒ y + x = xy
Dividing by xy
\(\frac{y}{x y}+\frac{x}{x y}=\frac{x y}{x y} \Rightarrow \frac{1}{x}+\frac{1}{y}\) = 1
Hence \(\frac { 1 }{ x }\) + \(\frac { 1 }{ y }\) = 1