# Mathematics Questions and Answers – Various Forms of the Equation of a Line

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This set of Mathematics Problems for Class 11 focuses on “Various Forms of the Equation of a Line”.

1. If slope of a line is positive then its inclination is ___________
a) right angle
b) acute angle
c) obtuse angle
d) zero

Explanation: If inclination is α slope is given by tan α. Given that slope of line is positive which means tan α is positive. We know, tan α is positive in 1st quadrant i.e. α should be acute angle.

2. If slope of a line is negative then its inclination is ___________
a) right angle
b) acute angle
c) obtuse angle
d) zero

Explanation: If inclination is α slope is given by tan α. Given that slope of line is negative which means tan α is negative. We know, tan α is negative in 2nd quadrant i.e. α should be obtuse angle.

3. Find slope of line joining (1, 2) and (4, 11).
a) 1/3
b) 3
c) 9
d) 1/9

Explanation: We know, slope of line joining two points (x1, y1) and (x2, y2) is given by $$\frac{y_2-y_1}{x_2-x_1}$$.
So, slope of line joining (1, 2) and (4, 11) is $$\frac{11-2}{4-1} = \frac{9}{3}$$ = 3.

4. If two lines are parallel their inclination angle may be different.
a) True
b) False

Explanation: If two lines are parallel then they form same angle with positive direction of x-axis in anticlockwise direction i.e. their inclinations are equal.

5. If two lines are parallel then their slopes must be equal.
a) True
b) False

Explanation: Let the inclination of the two lines be α and β. Since they are parallel so, α = β.
=>tan α = tan β. Hence their slopes are equal.

6. If the two lines are perpendicular then difference of their inclination angle is ________
a) 45°
b) 60°
c) 90°
d) 180°

7. If the two lines with slope m1 and m2 are perpendicular then their slopes has relation ______
a) m1 + m2 = 1
b) m1 * m2 = 1
c) m1 * m2 = -1
d) m1 + m2 = -1

Explanation: If the two lines are perpendicular then if one line form angle α with positive x-axis then the other line form angle 90° + α.
If m1 = tan α then m2 will be tan (90°+ α) = – cot α = -1/tan α
=> m1 * m2 = – 1.

8. If angle between the two lines is 45° and slope of one line is 1/4 then which of the following is possible value of the slope of other line.
a) 5/3
b) 3/5
c) -5/3
d) 4/5

Explanation: If angle between two lines with slopes m1 and m2 is α then tan α = |(m1-m2)/(1+m1*m2)|
tan 450 = $$|\frac{m-1/4}{1+m/4}| = \frac{4m-1}{m+4}$$
=>1 = $$\frac{4m-1}{m+4}$$ => m+4 = 4m-1 => 3m = 5
=>m = 5/3.

9. If angle between the two lines is 45° and slope of one line is 1/4 then which of the following is possible value of the slope of other line.
a) 3/5
b) -3/5
c) -5/3
d) 4/5

Explanation: If angle between two lines with slopes m1 and m2 is α then tan α = |(m1-m2)/(1+m1*m2)|
tan 45° = $$|\frac{m-1/4}{1+m/4}|$$
=> $$\frac{4m-1}{m+4}$$ = -1
=>- m-4 = 4m-1 => 5m = -3
=> m = -3/5.

10. If slope of a line is 2/3 then find the slope of line perpendicular to it.
a) -3/2
b) 3/2
c) 2/3
d) -2/3

Explanation: If lines with slopes m1 and m2 are perpendicular then m1 * m2 = – 1.
If m1 = 2/3 then m2 = -1 / (2/3) = -3/2.

11. If slope of one line is 1/4 and other is 5/3 then find the angle between two lines.
a) 30°
b) 45°
c) 90°
d) 180°

Explanation: If angle between two lines with slopes m1 and m2 is α then tan α = |(m1-m2)/(1+m1*m2)|
tan α = $$|\frac{5/3-1/4}{1+5/3*1/4}| = |\frac{20-3}{12+5}|$$ = 17/17 =1 => α = 45°

12. Find slope of line passing through origin and (3, 6).
a) 2
b) 3
c) 1/3
d) 1/2

Explanation: We know, slope of line joining two points (x1, y1) and (x2, y2) is given by (y2-y1)/(x2-x1).
So, slope of line joining (0, 0) and (3, 6) is (6-0)/(3-0) = 6/3 = 2.

13. If line joining (1, 2) and (5, 7) is parallel to line joining (3, 4) and (11, x).
a) 10
b) 11
c) 12
d) 14

Explanation: We know, slope of line joining two points (x1, y1) and (x2, y2) is given by(y2-y1)/(x2-x1).
Lines are parallel means slope is equal.
=>(x-4)/(11-3) = (7-2)/(5-1) => x-4 = 5*8/4 = 10 => x=14.

14. If line joining (1, 2) and (7, 6) is perpendicular to line joining (3, 4) and (11, x).
a) 12
b) 16
c) -16
d) -12

Explanation: We know, slope of line joining two points (x1, y1) and (x2, y2) is given by(y2-y1)/(x2-x1).
Lines are perpendicular means m1*m2 = -1
=> $$(\frac{x-4}{11-3})(\frac{6-2}{7-1})$$ = -1
=> (x-4)(4) = (-1)(8)(6)
=> x-4 = -12 => x= -16.

15. The points A (1, 2), B (3, 5), C (7, 8) are collinear.
a) True
b) False

Explanation: We know, slope of line joining two points (x1, y1) and (x2, y2) is given by(y2-y1)/(x2-x1).
Slope of line AB = (5-2)/(3-1) = 3/2
Slope of line BC = (8-5)/(7-3) = 3/4
Since slope of AB is not equal to BC so, points are not collinear.

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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