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

View Answer

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

View Answer

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

View Answer

Explanation: Given equation is y=5x+10. Y-intercept means value of y when x is zero

y=0+10 => y=10

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

View Answer

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

View Answer

Explanation: Given equation is x + y=2. Reducing the above equation to normal form

(x + y)/√2 =√2.

x cos45° + 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°

View Answer

Explanation: Given equation is x + y=2. Reducing the above equation to normal form

(x + y)/√2 =√2.

x cos45° + 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

View Answer

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

View Answer

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) \(\frac{3}{\sqrt{2}}\)

View Answer

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) \(\frac{3}{\sqrt{2}}\)

View Answer

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 \(|\frac{8-4}{\sqrt{2^2+1^2}}|=\frac{4}{\sqrt{5}}\).

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