# Class 11 Maths MCQ – General Equation of a Line

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

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

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

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

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

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

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

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

Explanation: We know, equation of line joining two points (x1, y1) and (x2, y2) 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

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

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

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

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 cos⁡ω + y sin⁡ω = p.
So, equation of given line will be x cos⁡45° + y sin⁡45° = 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

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

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

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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