OP Malhotra Class 11 Maths Solutions Chapter 16 The Straight Line Ex 16(a)

Utilizing Class 11 ISC Maths S Chand Solutions Chapter 16 The Straight Line Ex 16(a) as a study aid can enhance exam preparation.

S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(a)

Question 1.
Find the slope of a line whose inclination is
(i) 30°
(ii) 45°
(iii) 60°
(iv) 15°
(v) 135°
Solution:
(i) Given inclination of line = θ = 30°
∴ slope of line = tan 30° = \(\frac{1}{\sqrt{3}}\)

(ii) Given inclination of line θ = 45°
∴ slope of line = tan θ = tan 45° = 1

(iii) Given inclination of line = θ = 60°
∴ slope of line = tan θ = tan 60° = \(\sqrt{3}\)

(iv) Given inclination of line θ = 15°
∴ required slope of line = tan 15° = tan (45° – 30°)

(v) Given inclination of line θ = 135°
∴ required slope of line = tan 135°
= tan (180° – 45°)
= – tan 45° = – 1

Question 2.
Find the slope and inclination of the line through each pair of the following points :
(i) (1, 2) and (5, 6)
(ii) (0, 0) and (-√3, 3)
(iii) (10, 4) and (- 2, – 2)
(iv) (- 1, – 8) and (5, 7).
Solution:
We know that slope of line joining two points (x₁, y₁) and (x₂, y₂) = m = \(\frac{y_2-y_1}{x_2-x_1}\)

(i) ∴ Slope of line joining two points (1, 2) and (5, 6) = \(\frac{6-2}{5-1}\) = 1 = m
Let θ be the inclination of line Then tan θ = m = 1 ⇒ θ = 45°

(ii) Slope of line joining the points (0, 0) and (-√3, 3)

Let θ be the inclination of the line Then m = tan θ
∴ tan θ = – √3 = – tan\(\frac{\pi}{3}\) = tan \(\left(\pi-\frac{\pi}{3}\right)\)
⇒ θ = π – \(\frac{\pi}{3}\) = \(\frac{2 \pi}{3}\) or 120°

(iii) Thus, slope of line joining the points (10, 4) and (- 2, – 2) = m
= \(\frac{-2-4}{-2-10}\) = \(\frac{-6}{-12}\) = \(\frac{1}{2}\)

Let θ be the inclination of line then slope of line = m = tan θ
∴ tan θ = \(\frac{1}{2}\) ⇒ θ = 26°34′

(iv) ∴ slope of line joining the points (- 1, – 8) and (5, 7) be m = \(\frac{7+8}{5+1}\) = \(\frac{15}{6}\) = \(\frac{5}{2}\)
Let θ be the inclination of the line Then slope of the line = tan θ
∴ tan θ = \(\frac{5}{2}\) ⇒ θ = 68°12′

Question 3.
In the hexagon PQRSTU, RS || P U || QT. Which sides or diagonals have
(i) positive slope
(ii) negative slope
(iii) zero slope
(iv) infinite slope ?

Solution:
(i) If θ be the inclination of line. Then slope of line be positive if θ > 0
i.e. 0 < θ < 90°.
Then the lines QR, QS, PS, PT, PR are having positive slope.

(ii) Then line with negative slope means tan θ < 0 ⇒ 90° < θ < 180°.
Clearly the lines RT, QU, RU, ST and SU are having negative slopes.

(iii) The line is having zero slope
∴ tan θ = 0 ⇒ θ = 0
Thus the lines PU, QT and RS are having 0 slope.

(iv) Clearly the line is having infinite slope
∴ tan θ → ∞ ⇒ θ → \(\frac{\pi}{2}\)
Thus the lines PQ and UT are having infinite slopes.

Question 4.
The side BC of an equilateral △ABC is parallel to the x-axis. What are the slopes of its sides?
Solution:
Since the side BC of equilateral △ABC is parallel to x-axis.
∴ inclination of side BC = 0°
Thus slope of side BC = tan 0° = 0

Clearly the side AC makes an angle of 120° with +ve direction of x-axis.
∴ slope of side AC = tan 120°
= tan (180° – 60°)
= – tan 60° = -√3
Clearly the side AB makes an angle of 60° with positive direction of x-axis.
∴ slope of side AB of △ABC = tan 60° = +√3

Question 5.
In a regular hexagon ABCDEF, AB || E D || the x-axis. What are the slopes of its sides?
Solution:
Let the side AB and AE of hexagon ABCDEF along x-axis and y-axis.

We know that, exterior angle of hexagon = \(\frac{360^{\circ}}{6}\) = 60°
∴ inclinations of sides BC, CD, DE, EF and AF are 60°, 120°, 0°, 60° and 120°
Thus slopes of sides BC, CD, DE, EF and AF are tan 60°, tan 120°, tan 0°, tan 60° and tan (180° – 60°)
i.e. √3, – √3, 0, √3, – √3.

Question 6.
Using slopes determine which of the following sets of three points are collinear.
(i) (5, – 2), (7, 6), (0, – 2);
(ii) (- 2, 3), (8, – 5), (5, 3);
(iii) (6, – 1), (5, 0), (2, 3);
(iv) (-1, 5), (3, 1), (5, 7).
Solution:
(i) Given points are A(5, – 2); B(7, 6) and C(0, – 2)
slope of line AB = \(\frac{6+2}{7-5}\) = 4

slope of line BC = \(\frac{-2-6}{0 – 7}\) = \(\frac{8}{7}\)
and slope of line AC = \(\frac{-2+2}{0 – 5}\) = 0
∴ Slope of line AB ≠ slope of BC ≠ slope of AC
Thus, given points A, B and C are not collinear.

(ii) Given points are A (- 2, 3), B(8, – 5) and C (5, 4).
∴ slope of line AB = \(\frac{-5-3}{8 + 2}\) = \(\frac{-8}{10}\) = \(\frac{-4}{5}\)
slope of line BC = \(\frac{4+5}{5-8}\) = \(\frac{9}{-3}\) = -3
and slope of line AC = \(\frac{4-3}{5+2}\) = \(\frac{1}{7}\)
Clearly points A, B and C are not collinear.

(iii) Given points are A(6,- 1), B(5, 0) and C(2, 3).
∴ slope of line AB = \(\frac{0+1}{5-6}\) = – 1
slope of line BC = \(\frac{3-0}{2-5}\) = – 1
and slope of line AC = \(\frac{3+1}{2-6}\) = – 1
∴ slope of AB = slope of BC = slope of AC
Thus the points A, B and C are collinear.

(iv) Given points are A(- 1, 5), B(3, 1) and C(5, 7).
∴ slope of line AB = \(\frac{1-5}{3+1}\) = – 1
slope of line BC = \(\frac{7-1}{5-3}\) = \(\frac{6}{2}\) = 3
and slope of line AC = \(\frac{7-5}{5+1}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
Clearly the given points A, B and C are not collinear.

Question 7.
Find y if the slope of the line joining (- 8, 11),(2, y) is \(\frac{-4}{3}\).
Solution:
slope of line joining the points (- 8, 11) and (2, y) = \(\frac{y-11}{2+8}\) = \(\frac{y-11}{10}\)
Also given slope of line be –\(\frac{4}{3}\).
∴ \(\frac{y-11}{10}\) = –\(\frac{4}{3}\) ⇒ 3y – 33 = – 40
⇒ 3y = 33 – 40 ⇒ y = –\(\frac{7}{3}\)

Question 8.
Find the angle between the lines whose slope are (i) 2 and -1 (ii) 2 and \(\frac{3}{4}\).
Solution:
(i) Given m₁ = 2 and m₂ = – 1
Let θ be the acute angle between the lines

Question 9.
Find the slope of the line which makes an angle of 45° with a line of slope \(\frac{-6}{5}\).
Solution:
Let m be the slope of line which makes an angle of 45° with a line of slope \(\frac{-6}{5}\)

Hence the slopes of required line be 11 and \(\frac{-1}{11}\).

Question 10.
Find the interior angles of the triangles whose vertices are (4, 3) B (-2, 2) and C (2, – 8).
Solution:
slope of line AC = \(\frac{-8-3}{2-4}\) = \(\frac{11}{2}\)
slope of side BC = \(\frac{-8-2}{2+2}\) = \(\frac{-5}{2}\)
slope of side AB = \(\frac{3-2}{4+2}\) = \(\frac{1}{6}\)

We know that, if θ be the angle between the lines having slopes m₁ and m₂.

Question 11.
Find the slope of a line parallel to a line whose slope is
(i) -3
(ii) \(\frac{1}{2}\)
(iii) 2.3
(iv) 0
Solution:
Two lines are having slopes m₁ and m₂ are parallel iff m₁ = m₂
(i) Thus slope of line which is || to line having slope – 3 = – 3
(ii) Slope of line which is parallel to line having slope \(\frac{1}{2}\) = \(\frac{1}{2}\)
(iii) Slope of line which is parallel to line having slope 2.3 = 2.3
(iv) Slope of line which is || to line having slope 0 = 0

Question 12.
Find the slope of a line parallel to the line which passes through each pair of the following points :
(i) (0, 0) and (5, 6),
(ii) (- 1, 3) and (4, 7),
(iii) (- 5, – 8) and (3, 0),
(iv) (-a, 0) and (0, b).
Solution:
(i) Slope of line passes through the points (0, 0) and (5, 6)

∴ slope of line parallel to line joining (0, 0) and (5, 6) = \(\frac{6}{5}\)
[If two lines having slopes m₁ and m₂ are parallel. Then m₁ = m₂ ]

(ii) Slope of line joining (-1, 3) and (4, 7) = \(\frac{7-3}{4-(-1)}\) = \(\frac{4}{5}\)
∴ slope of line parallel to line joining (-1, 3) and (4, 7) = \(\frac{4}{5}\)
[∵ slopes of parallel lines are equal]

(iii) Slope of line passes through the points
(-5, – 8) and (3, 0) = \(\frac{0+8}{3+5}\) = \(\frac{8}{8}\) = 1
∴ slope of line parallel to line passes through the point (- 5, – 8) and (- 3, 0) be 1 .

(iv) slope of line joining the points (-a, 0) and (0, b) = \(\frac{b-0}{0+a}\) = \(\frac{b}{a}\)
Thus slope of line parallel to line passes through the point (-a, 0) and (0, b) = \(\frac{b}{a}\)

Question 13.
Find the slope of a line perpendicular to the line whose slope is
(i) \(\frac{1}{3}\)
(ii) –\(\frac{5}{6}\)
(iii) 5
(iv) -5\(\frac{1}{7}\)
(v) 0
(vi) Infinite
Solution:
We know that two lines are perpendicular if product of their slopes is -1
∴ m₁ m₂ = -1

(i) given slope of line = \(\frac{1}{3}\)
∴ slope of line ⊥ to given line = –\(\frac{1}{\frac{1}{3}}\) = -3

(ii) Slope of given line = –\(\frac{5}{6}\)
∴ slope of line ⊥ to given line = \(\frac{-1}{\frac{-5}{6}}\) = \(\frac{6}{5}\)

(iii) Slope of given line be 5
∴ slope of line ⊥ to given line be \(\frac{-1}{5}\).

(iv) Slope of given line = -5\(\frac{1}{7}\) = –\(\frac{36}{7}\)
∴ slope of line ⊥ to given line = \(\frac{-1}{\frac{-36}{7}}\) = \(\frac{7}{36}\)

(v) Slope of given line = 0
∴ Slope of line ⊥ to given line = \(\frac{1}{0}\) = + ∞

(vi) Slope of given line = ∞
∴ Slope of line ⊥ to given line = \(\frac{1}{\infty}\) = 0

Question 14.
Find the slope of a line perpendicular to the line which passes through each pair of the following points :
(i) (0, 8) and (-5, 2);
(ii) (1, – 11) and (5, 2);
(iii) (-k, h) and (b,- f)
(iv) (x₁, y₁) and (x₂, y₂)
Solution:
(i) Slope of line passes through the points (0, 8) and (-5, 2) = \(\frac{2-8}{-5-0}\) = \(\frac{-6}{-5}\) = \(\frac{6}{5}\)
∴ slope of line ⊥ to given line \(\frac{\frac{-1}{-6}}{5}\) = –\(\frac{5}{6}\) [∵ m₁ m₂ = -1]

(ii) Slope of line passes through the points (1, – 11) and (5, 2)
\(\frac{2+11}{5-1}\) = \(\frac{13}{4}\) [∵ slope = \(\frac{y_2-y_1}{x_2-x_1}\)]
Thus slope of line ⊥ to the line passes through the points (1, – 11) and (5, 2)
= \(\frac{-1}{\frac{13}{4}}\) = \(\frac{-4}{13}\) [∵ m₁m₂ = -1]

(iii) Slope of line passes through the points (- k, h) and (b, – f)

(iv) Slope of line joining the points(x₁, y₁)
and (x₂, y₂) = \(\frac{y_2-y_1}{x_2-x_1}\)
∴ slope of line ⊥ to given line = \(-\frac{1}{\frac{y_2-y_1}{x_2-x_1}}\) = \(-\left(\frac{x_2-x_1}{y_2-y_1}\right)\)

Question 15.
In rect. ABCD, slope of AB = \(\frac{5}{6}\). State the slope of (i) BC, (ii) CD, (iii) DA.
Solution:
Given slope of AB = \(\frac{5}{6}\)
Since ABCD is a rectangle.
∴ AB || DC and AD || BC
Further AB ⊥ BC
since AB || DC
∴ slope of CD = slope of AB = \(\frac{5}{6}\)

∴ slope of BC = –\(\frac{1}{\text { slope of } A B}\) = \(\frac{-1}{\frac{5}{6}}\) = –\(\frac{6}{5}\)
Also DA || BC
∴ slope of DA = slope of BC = –\(\frac{6}{5}\)

Question 16.
In Parallelogram ABCD,
slope of AB = -2, slope of BC = \(\frac{3}{5}\).
state the slope of
(i) AD
(ii) CD
(iii) the altitude of AD,
(iv) the altitude of CD.
Solution:
Given slope of AB = – 2
and slope of BC = \(\frac{3}{5}\)
since ABCD is a parallelogram
∴ AB || DC.

∴ slope of DC = slope of AB = – 2
slope of AD = slope of BC = \(\frac{3}{5}\) [∵ AD || BC]
Clearly slope of the altitude to AD
= – \(\frac{1}{\text { slope of } A D}\) =\(\frac{-1}{\frac{3}{5}}\) = \(\frac{-5}{3}\)
Clearly slope of the altitude to CD
= \(\frac{-1}{\text { slope of line } B C}\) = \(\frac{-1}{-2}\) = \(\frac{1}{2}\)

Question 17.
The vertices of a △ABC are A(1, 1), B(7, 3) and C(3, 6). State the slope of the altitude to
(i) AB,
(ii) BC,
(iii) AC.
Solution:
slope of AB = \(\frac{3-1}{7-1}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
slope of BC = \(\frac{6-3}{3-7}\) = \(-\frac{3}{4}\)
slope of AC = \(\frac{6-1}{3-1}\) = \(\frac{5}{2}\)

Question 18.
The line joining (-5, 7) and (0, – 2) is perpendicular to the line joining (1, – 3) and (4, x). Find x.
Solution:
The slope of line joining (- 5, – 7) and (0, – 2) = m₁ = \(\frac{-2-7}{0+5}\) = \(\frac{-9}{5}\)
and slope of line joining (1, – 3) and (4, x)
=m₂ = \(\frac{x+3}{4-1}\) = \(\frac{x+3}{3}\)
Since the line joining (- 5, 7) and (0, – 2) is ⊥ to the line joining (1, – 3) and (4, x).
∴ m₁ m₂ = – 1
⇒ \(\left(\frac{-9}{5}\right)\) \(\left(\frac{x+3}{3}\right)\) = – 1
⇒ -9x – 27 = – 15
⇒ -9x = 12
⇒ x = \(\frac{-12}{9}\) = \(\frac{-4}{3}\)