This set of Class 11 Maths Chapter 10 Multiple Choice Questions & Answers (MCQs) focuses on “General Equation of a Line”.

1. Equation of horizontal line above x-axis at 5 units from x-axis is _______

a) x=5

b) x=-5

c) y=5

d) y=-5

View Answer

Explanation: Equation of x-axis is y=0. Horizontal line is parallel to x-axis and above it by 5 units so, equation of line is y=5.

2. Equation of horizontal line below x-axis at 5 units from x-axis is _______

a) x=5

b) x=-5

c) y=5

d) y=-5

View Answer

Explanation: Equation of x-axis is y=0. Horizontal line is parallel to x-axis and below it by 5 units so, equation of line is y=-5.

3. Equation of vertical line to the right of y-axis at 5 units from y-axis is _______

a) x=5

b) x=-5

c) y=5

d) y=-5

View Answer

Explanation: Equation of y-axis is x=0. Vertical line is parallel to y-axis and to the right by 5 units so, equation of line is x=5.

4. Equation of vertical line to the left of y-axis at 5 units from y-axis is _______

a) x=5

b) x=-5

c) y=5

d) y=-5

View Answer

Explanation: Equation of y-axis is x=0. Vertical line is parallel to y-axis and to the left by 5 units so, equation of line is x=-5.

5. Find the equation of line parallel to x-axis and passing through (3, 4).

a) x=3

b) x=4

c) y=4

d) y=3

View Answer

Explanation: Let general equation of line be y=m*x + c.

Since line is parallel to x-axis so, m=0.

=>y=c=>y=4 by substituting the point (3, 4).

6. Find the equation of line parallel to y-axis and passing through (3, 4).

a) x=3

b) x=4

c) y=4

d) y=3

View Answer

Explanation: Let general equation of line be y=m (x-d) => x = y/m +d

Since line is parallel to y-axis so, m=1/0 or 1/m =0

=>x=d => x=3 by substituting the point (3, 4).

7. If a line has slope 3 and pass through point (1, 2) then the equation of line is _________

a) x=3y-1

b) x=3y+1

c) y=3x+1

d) y=3x-1

View Answer

Explanation: Let general equation of line be y=m*x + c.

Given m=3 => y=3x+c

Substituting the point (1, 2) in above equation we get 2=3*1 + c => c = -1

So, equation of line will be y = 3x-1.

8. Find the equation of line joining (3, 4) and (5, 8).

a) y = 2x-1

b) y = 2x+2

c) y = 2x+1

d) y = 2x-2

View Answer

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

_{1}, y

_{1}) and (x

_{2}, y

_{2}) is given by

\(\frac{y-y1}{y2-y1} = \frac{x-x1}{x2-x1}\)

So, equation will be \(\frac{y-4}{8-4} = \frac{x-3}{5-3}\)

=>\(\frac{y-4}{4} = \frac{x-3}{2}\)

=>y-4 = 2x-6

=>y = 2x-2

9. If slope of a line is 4 and y-intercept made by the line is 2 then the equation of line will be __________

a) y=4x-2

b) y=4x+2

c) y=2x+4

d) y=2x-4

View Answer

Explanation: Let general equation of line be y=m*x + c.

Given, m=4 and c=2.

=>y=4x+2

10. If slope of a line is 4 and x-intercept made by the line is 2 then the equation of line will be __________

a) y = 4x-8

b) y = 4x+8

c) y = 2x+4

d) y = 2x-4

View Answer

Explanation: Let general equation of line be y=m*x + c.

Given m=4 and value of x when y=0 is 2.

C= – m*2 = -4*2 = -8.

=>y=4x-8.

11. If x-intercept of a line is 4 and its y-intercept is 2 then find the equation of line.

a) 2x+y-4=0

b) x+2y-4=0

c) 2x+y+4=0

d) x+2y+4=0

View Answer

Explanation: If x-intercept of a line is a and y-intercept of line is b so, equation of line is \(\frac{x}{a} + \frac{y}{b}\) = 1.

Equation of line is \(\frac{x}{4} + \frac{y}{2}\) = 1 => x+2y-4=0.

12. If perpendicular distance of a line from origin is 4 units and angle which the normal makes with positive x-axis is 45°, then the equation will be ______________

a) x + y = 4√2

b) x – y = 4√2

c) y – x = 4√2

d) x + y = -4√2

View Answer

Explanation: If p is perpendicular distance of line from origin and ω be the angle formed by normal with positive x-axis then equation of line is x cosω + y sinω = p.

So, equation of given line will be x cos45° + y sin45° = 4

=>(x + y)/√2 = 4

=>x + y = 4√2.

13. If -40°F is equal to -40°C and 0°C is equal to 32°F then find the value of 40°C.

a) 104°F

b) 112°F

c) 86°F

d) 92°F

View Answer

Explanation: Let general equation be F=m*C +k

-40 = -40m + k

and 32 = 0 + k

=>-40 = -40m + 32 => m=72/40 = 18/10

F=18/10 * 40 + 32 = 72+32 = 104.

14. If P (1, 2), Q (3, 5), R (7, 9) form a triangle then find the equation of median through P.

a) 5x-4y+3 = 0

b) 5x+4y+3 = 0

c) 5x-4y-3 = 0

d) 5x+4y-3 = 0

View Answer

Explanation: Midpoint of QR line is \((\frac{3+7}{2}, \frac{5+9}{2})\) = (5, 7).

Equation of line joining (1, 2) and (5, 7) is \(\frac{y-2}{7-2} = \frac{x-1}{5-1}\)

=>\(\frac{y-2}{5} = \frac{x-1}{4}\) => 4y-8 = 5x-5 => 5x-4y+3 =0.

15. Line perpendicular to line joining (1, 3) and (4, 5) and pass through (2, 4) has equation ____________

a) 3x+2y-14=0

b) 3x-2y-14=0

c) 2x-3y-14=0

d) 2x+3y-14=0

View Answer

Explanation: Slope of line joining (1, 3) and (4, 5) is (5-3) / (4-1) = 2/3

So, slope of line perpendicular to it is -3/2.

Line has slope -3/2 and pass through (2, 4)

So, equation is y-4 = (-3/2)(x-2) => 2y-8 = -3x+6 => 3x+2y-14=0.

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

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