Mathematics Questions and Answers – Distance of a Point from a Line

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Distance of a Point from a Line”.

1. If equation of a line is y=3x-4 then find the slope of line.
a) 3
b) -3
c) 4
d) -4

Explanation: Comparing the above equation with general equation y=m*x + c,
m=3 which is the slope of line.

2. If equation of line is y=5x+10 then find the value of x-intercept made by the line.
a) 2
b) 1/2
c) -1/2
d) -2

Explanation: Given equation is y=5x+10. X-intercept means value of x when y is zero
0=5x+10 => x=-2

3. If equation of line is y=5x+10 then find the value of x-intercept made by the line.
a) 10
b) 1/10
c) -1/10
d) -10

Explanation: Given equation is y=5x+10. Y-intercept means value of y when x is zero
y=0+10 => y=10
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4. If equation of a line is 3x+2y-6=0 then x-intercept is _____ and y-intercept is _____
a) 3, 2
b) 2, 3
c) 2, 6
d) 3, 6

Explanation: Reducing the above equation to intercept form x/a +y/b =1,
we get x/2 +y/3 =1
a=2 which is x-intercept and b=3 which is y-intercept.

5. If equation of line is x + y=2 then find the perpendicular distance of line from origin.
a) √2
b) √3
c) 1/√2
d) 1/√3

Explanation: Given equation is x + y=2. Reducing the above equation to normal form
(x + y)/√2 =√2.
x cos⁡45° + y sin 45° = √2
Perpendicular distance from origin is √2.

6. If equation of line is x + y=2 then find the angle made by line with x-axis.
a) 45°
b) 60°
c) 30°
d) 75°

Explanation: Given equation is x + y=2. Reducing the above equation to normal form
(x + y)/√2 =√2.
x cos⁡45° + y sin 45° = √2
Angle made with x-axis is 45°.

7. Find the equation perpendicular to 2x – y=4 and pass through (2, 4).
a) 2x+y-10 = 0
b) x+2y+10 = 0
c) x+2y-10 = 0
d) x+y-10 = 0

Explanation: Line 2x – y=4 has slope 2. Line perpendicular to given line has slope -1/2.
Equation is (y-4) = (-1/2) (x-2)
2y-8 = -x + 2 => x+2y -10 = 0.

8. Find the equation of line parallel to 4x+y=2 and pass through (2, 5).
a) 4x+y-13=0
b) 4x+y+13=0
c) 4x-y-13=0
d) 4x-y+13=0

Explanation: Line 4x+y=2 has slope -4. Line parallel to it has slope -4 and pass through (2, 5) so equation will be y-5 = (-4) (x-2) => 4x+y-13=0

9. Find the perpendicular distance of point (2, 4) from line 2x+y-5=0.
a) $$\frac{6}{\sqrt{5}}$$
b) $$\frac{3}{\sqrt{5}}$$
c) $$\frac{9}{\sqrt{5}}$$
d) $$\frac{3}{\sqrt{2}}$$

Explanation: Distance of point (x1, y1) from line ax +by +c=0 is $$|\frac{ax1+by1+c}{\sqrt{a^2+b^2}}|$$.
So, distance of point (2, 4) from line 2x+y-5=0 is $$|\frac{2*2+4*1-5}{\sqrt{2^2+1^2}}| = \frac{3}{\sqrt{5}}$$.

10. Find the distance between 2x+y+4=0 and 2x+y+8=0.
a) $$\frac{4}{\sqrt{5}}$$
b) $$\frac{3}{\sqrt{5}}$$
c) $$\frac{9}{\sqrt{5}}$$
d) $$\frac{3}{\sqrt{2}}$$

Explanation: Distance between parallel lines ax +by +c1=0 and ax+by+c2=0 is $$|\frac{c1-c2}{\sqrt{a^2+b^2}}|$$.
So, distance 2x + y+4=0 and 2x+y+8=0 is $$|\frac{8-4}{\sqrt{2^2+1^2}}|=\frac{4}{\sqrt{5}}$$.

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

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