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

View Answer

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 1

^{st}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

View Answer

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 2

^{nd}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

View Answer

Explanation: We know, slope of line joining two points (x

_{1}, y

_{1}) and (x

_{2}, y

_{2}) 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

View Answer

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

View Answer

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°

View Answer

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.

7. If the two lines with slope m_{1} and m_{2} are perpendicular then their slopes has relation ______

a) m_{1} + m_{2} = 1

b) m_{1} * m_{2} = 1

c) m_{1} * m_{2} = -1

d) m_{1} + m_{2} = -1

View Answer

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 m

_{1}= tan α then m

_{2}will be tan (90°+ α) = – cot α = -1/tan α

=> m

_{1}* m

_{2}= – 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

View Answer

Explanation: If angle between two lines with slopes m

_{1}and m

_{2}is α then tan α = |(m

_{1}-m

_{2})/(1+m

_{1}*m

_{2})|

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

View Answer

Explanation: If angle between two lines with slopes m

_{1}and m

_{2}is α then tan α = |(m

_{1}-m

_{2})/(1+m

_{1}*m

_{2})|

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

View Answer

Explanation: If lines with slopes m

_{1}and m

_{2}are perpendicular then m

_{1}* m

_{2}= – 1.

If m

_{1}= 2/3 then m

_{2}= -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°

View Answer

Explanation: If angle between two lines with slopes m

_{1}and m

_{2}is α then tan α = |(m

_{1}-m

_{2})/(1+m

_{1}*m

_{2})|

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

View Answer

Explanation: We know, slope of line joining two points (x

_{1}, y

_{1}) and (x

_{2}, y

_{2}) is given by (y

_{2}-y

_{1})/(x

_{2}-x

_{1}).

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

View Answer

Explanation: We know, slope of line joining two points (x

_{1}, y

_{1}) and (x

_{2}, y

_{2}) is given by(y

_{2}-y

_{1})/(x

_{2}-x

_{1}).

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

View Answer

Explanation: We know, slope of line joining two points (x

_{1}, y

_{1}) and (x

_{2}, y

_{2}) is given by(y

_{2}-y

_{1})/(x

_{2}-x

_{1}).

Lines are perpendicular means m

_{1}*m

_{2}= -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

View Answer

Explanation: We know, slope of line joining two points (x

_{1}, y

_{1}) and (x

_{2}, y

_{2}) is given by(y

_{2}-y

_{1})/(x

_{2}-x

_{1}).

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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