The point on the line 4 x - y - 2 = 0 which is equidistant from the points \left(\right. - 5 , 6 \left.\right) and \left(\right. 3 , 2 \left.\right) is

Engineering

Mathematics

Locus of a Point

Various Form of a Straight Line

Question

The point on the line

4 x - y - 2 = 0

which is equidistant from the points

(- 5, 6)

and

(3, 2)

(2, 6)

(4, 14)

(1, 2)

(3, 10)

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Let the point on line

4 x - y - 2 = 0

P (x, y)

Let

A \equiv (- 5, 6)

and

B \equiv (3, 2)

4 x - y - 2 = 0

....(1)

Point P is equidistant from points A and B ....Given

∴ A P = P B

By distance formula,

(x + 5)^{2} + (y - 6)^{2} = (x - 3)^{2} + (y - 2)^{2}

x^{2} + 10 x + 25 + y^{2} - 12 y + 36 = x^{2} - 6 x + 9 + y^{2} - 4 y + 4

16 x - 8 y + 48 = 0

4 x - 2 y + 12 = 0

....(2)

Subtract eq(2) from eq (1), we get

y = 14

Substitute in eq(1), we get

x = 4

So, the point on the line is

(4, 14)