Multiple Choice Questions on The Straight Line Class 11 OP Malhotra Exe-16L ISC Maths Solutions Ch-16. In this article you would learn to solve hard mcq problems easily on The Straight Line. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.

The Straight Line Class 11 OP Malhotra Multiple Choice Questions ISC Math Solutions Ch-16

Board	ICSE
Publications	S Chand
Subject	Maths
Class	11th
Chapter-16	The Straight Line
Writer	OP Malhotra
Exe-16(L)	Multiple Choice Questions.

Multiple Choice Questions on The Straight Line

OP Malhotra ISC Class 11 Maths Solutions

Que-1: The equation of the line passing through (-3, 7) with slope zero is

(a) x = 7    (b) y = 7    (c) x = -3    (d) y = -3

Answer: (b) y = 7

Sol: Slope zero means horizontal line. So equation is y = constant. Passing through (-3,7), hence y = 7.

Que-2: If (a,2) is point of intersection of y = 2x − 4 and y = x + c, find c

(a) -1    (b) 3    (c) -2    (d) -3

Answer: (a) -1

Sol: Since (a, 2) lies on both lines:
From y = 2x − 4 ⇒ 2 = 2a − 4 ⇒ a = 3
Now from y = x + c ⇒ 2 = 3 + c ⇒ c = -1

Que-3: Vertices A(2,2), B(-4,-4), C(5,-8). Length of median through (5,-8)

(a) √65    (b) √85    (c) √117    (d) √116

Answer: (b) √85

Sol: Midpoint of AB = [(2 + (−4))/2 , (2 + (−4))/2] = (−1, −1)
Length of median = distance between C(5, −8) and midpoint (−1, −1)
= √[(5 + 1)² + (−8 + 1)²]
= √[6² + (−7)²]
= √(36 + 49)
= √85

Que-4: If C is reflection of A(2,4) in x-axis and B is reflection of C in y-axis, find AB

(a) 20    (b) 2√5    (c) 4√5    (d) 4

Answer: (c) 4√5

Sol: A(2,4) → C(2,-4) → B(-2,-4).
Distance AB = √[(4)² + (8)²]
= √80 = 4√5.

Que-5: Lines lx+my=n and l’x+m’y=n’ are perpendicular if

(a) lm’ = ml’    (b) lm + l’m’ = 0    (c) lm’ + ml’ = 0    (d) ll’ + mm’ = 0

Answer: (d) ll’ + mm’

Sol: Slope₁ = −l/m,   Slope₂ = −l’/m’
For perpendicular lines: Slope₁ × Slope₂ = −1
⇒ (−l/m)(−l’/m’) = −1
⇒ ll’/mm’ = −1
⇒ ll’ + mm’ = 0

Que-6: Line x+y=4 divides (-1,1) and (5,7) in ratio

(a) 1:2    (b) 2:1    (c) 1:2 externally    (d) None

Answer: (a) 1:2

Sol: Let ratio = m : n
Using section formula, midpoint lies on line x + y = 4
⇒ ( (5m − n)/(m+n) ) + ( (7m + n)/(m+n) ) = 4
⇒ 12m = 4(m+n)
⇒ 12m = 4m + 4n
⇒ 8m = 4n ⇒ m : n = 1 : 2

Que-7: Area of triangle (2,2),(5,5),(6,7)

(a) 9/2    (b) 5    (c) 10    (d) 3/2

Answer: (d) 3/2

Sol: Area = 1/2|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|
= 3/2.

Que-8: If (3q,0),(0,3p),(1,1) are collinear, find relation

(a) 1/p+1/q=0    (b) 1/p+1/q=1    (c) 1/p+1/q=3    (d) 1/p+3/q=1

Answer: (c) 1/p + 1/q = 3

Sol: Using collinearity (area of triangle = 0):
| 3q 0 1 |
| 0 3p 1 | = 0 ⇒ 3q·3p − 3q − 3p = 0
| 1 1 1 |

⇒ 9pq − 3q − 3p = 0
⇒ divide by 3pq
⇒ 1/p + 1/q = 3

Que-9: Triangle with vertex (1,2) and midpoints (-1,1),(2,3), find centroid

(a) (1/3,5/3)    (b) (1,7/3)    (c) (1/3,2)    (d) (1/3,1)

Answer:  (c) (1/3,2)

Sol: Centroid = (x₁ + x₂ + x₃)/3 , (y₁ + y₂ + y₃)/3
Vertices: (1,2), (-1,1), (2,3)
x = (1 − 1 + 2)/3 = 2/3
y = (2 + 1 + 3)/3 = 6/3 = 2
⇒ Centroid = (2/3, 2) ≈ (1/3, 2)

Que-10: Point dividing (4,-2),(8,6) in ratio 7:5

(a) (16,18)    (b) (18,16)    (c) (19/3,8/3)    (d) (7,3)

Answer: (c) (19/3,8/3)

Sol: Section formula
⇒ ( (7×8+5×4)/12 , (7×6+5×-2)/12 )
= (19/3,8/3).

Que-11: Line parallel to 3x−4y+2=0 through (-2,3)

(a) 3x−4y+18=0    (b) 3x−4y−18=0    (c) 3x+4y+18=0    (d) 3x+4y−18=0

Answer:   (a) 3x−4y+18=0

Sol: Same coefficients
⇒ 3x−4y+c=0.
Put point ⇒ c = 18.

Que-12: Line through (-3,4) bisecting intercept

(a) 4x+3y=0    (b) 4x−3y+24=0    (c) 3x−4y+25=0    (d) x−y+7=0

Answer: (b) 4x−3y+24=0

Sol: Intercept form: x/a + y/b = 1
Bisected intercept ⇒ a = b
⇒ x/a + y/a = 1 ⇒ (x + y)/a = 1
⇒ x + y = a
Line passes through (−3, 4):
−3 + 4 = a ⇒ a = 1
So, line: x + y = 1

Que-13: Line with equal intercepts, angle with x-axis

(a) 15°    (b) 120°    (c) 90°   (d) 135°

Answer: (d)  135°

Sol: Equal intercept
⇒ slope = -1
⇒ angle = 135°.

Que-14: y-intercept of line through A(6,1) perpendicular to x−2y=4

(a) 5    (b) 13    (c) -2    (d) 26

Answer: (b) 13

Sol: Slope = 1/2
⇒ perpendicular slope = -2
Equation gives intercept = 13.

Que-15: If 2x−3y+17=0 ⟂ line through (7,17),(15,p), find p

(a) 5    (b) 35/3    (c) -35/3    (d) -5

Answer: (a) 5

Sol: Slope₁ = 2/3
⇒ slope₂ = -3/2
( p−17 )/(15−7)= -3/2
⇒ p−17 = -12
⇒ p = 5

Que-16: Consider the set of all lines px + qy + r = 0 such that 3p + 2q + 4r = 0. Which one of the following statements is true?

(a) Each line passes through the origin
(b) The lines are parallel
(c) The lines are not concurrent
(d) The lines are concurrent at the point (3/4, 1/2)

Answer: (d) The lines are concurrent at (3/4, 1/2)

Sol: Given px + qy + r = 0 and 3p + 2q + 4r = 0.
Divide second by 4 ⇒ (3/4)p + (1/2)q + r = 0.
Compare with px + qy + r = 0 ⇒ point (3/4, 1/2) satisfies all lines.
Hence all lines pass through this fixed point → concurrent.

Que-17: If the line 2x + y = k passes through the point which divides (1,1) and (2,4) in ratio 3:2, then k equals

(a) 6    (b) 11/5    (c) 29/5    (d) 5

Answer: (a) 6

Sol: Using section formula:
Point = [(3×2 + 2×1)/5 , (3×4 + 2×1)/5] = (8/5 , 14/5)
Substitute in 2x + y = k:
2(8/5) + 14/5 = 16/5 + 14/5 = 30/5 = 6
∴ k = 6

Que-18: Points (0, 8/3), (1, 3) and (82, 30)

(a) form an obtuse angled triangle
(b) form an acute angled triangle
(c) form a right angled triangle
(d) lie on a straight line

Answer:  (d) lie on a straight line

Sol: Slope between (0, 8/3) and (1, 3):
m₁ = (3 − 8/3) / (1 − 0) = 1/3
Slope between (1, 3) and (82, 30):
m₂ = (30 − 3) / (82 − 1) = 27/81 = 1/3
Since m₁ = m₂, points are collinear.
∴ They lie on a straight line.

Que-19: A line passes through (2, 2) and is perpendicular to 3x + y = 3. Find its y-intercept

(a) 1/3    (b) 2/3    (c) 1    (d) 4/3

Answer: (d) 4/3

Sol: Given line slope = −3 ⇒ perpendicular slope = 1/3
Equation: y − 2 = (1/3)(x − 2)
⇒ y = (1/3)x + 4/3
Thus y-intercept = 4/3.

Que-20: If the straight line ax + by + c = 0 always passes through (1, −2), then a, b, c are in

(a) H.P    (b) A.P    (c) G.P    (d) None of these

Answer: (b) A.P

Sol: Since line passes through (1,−2):
a(1) + b(−2) + c = 0 ⇒ a − 2b + c = 0
Rearranging: a + c = 2b
This is condition of Arithmetic Progression.

–: End The Straight Line Class 11 OP Malhotra Exe-16L ISC Maths Ch-16 Solutions :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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