# Distance of a Point From a Line

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• Q1

The distance between the parallel lines y = 2x + 4
and
6x = 3y + 5 is

Marks:1

##### Explanation:

• Q2

Let the algebraic sum of the perpendicular distances from the points (2, 0), (0, 2), (1, 1 ) to a variable line be zero, then the line passes through a fixed point whose coordinates are

Marks:1

(1, 1).

##### Explanation:

• Q3

Two lines are given (x – 2y)2 + k (x – 2y) = 0. The value of k, so that the distance between them is 3, is

Marks:1

k = 35.

##### Explanation:

The lines are given by

(x – 2y)2 + k (x – 2y) = 0

(x - 2y)(x - 2y + k) = 0

That is x  – 2y = 0 and x – 2y + k = 0

These are parallel. The distance between the two lines

• Q4

The equation of a straight line, which passes through the point (a, 0) and whose perpendicular distance from the point (2a, 2a) is a, is

Marks:1

3x – 4y – 3a = 0 and x – a = 0.

##### Explanation:

Equation of line passing through (a, 0) is y = m(x – a)
mx - y - ma = 0                                            ...(i)
Its distance from the point (2a, 2a) is
|(2am - 2a - ma)/(m2 + 1)| = a (given)

(m - 2)2 = (m2 + 1)

m2 - 4m + 4 = m2 + 1

- 4m + 3 = 0

m = 3/4, .

[If a = 0 then one root of ax2 + bx + c = 0 is infinite]

This result must be taken into care otherwise one root will be lost.

The required equation of lines is, from (i)

3x – 4y – 3a = 0

and       y = (x – a)
x – a = 0.