Question

Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.

Answer 1

Let the slope of the required line PQ be m.

The equation of the line PQ, which passes through the point P(−1, 2) and has slope m, is

y – y₁ = m(x – x₁)

y – 2 = m(x + 1)

or mx – y + m + 2 = 0       ....…(i)

equation of line AB x+ y = 4

∴ y = 4 – x

Putting the value of y in equation (1),

mx – (4 – x) + m + 2 = 0

or (m + 1) x + m – 2 = 0

∴ x = `- ("m" - 2)/("m" + 1)`

Now y = 4 − x

= `4 + ("m" - 2)/("m" + 1)`

= `(4"m" + 4 + "m" - 2)/("m" + 1) = (5"m" + 2)/("m" + 1)`

Given: PQ = 3 or PQ² = 9

∴ `(- ("m" - 2)/("m" + 1) + 1)^2 + ((5"m" + 2)/("m" + 1) - 2)^2 = 9`

or `((-"m" + 2 + "m" + 1)/("m" + 1))^2 + ((5"m" + 2 - 2"m" - 2)/("m" + 1))^2 = 9`

or `9/("m" + 1)^2 + ((3"m")/("m"+ 1))^2 = 9`

or `(9 + 9"m"^2)/("m" + 1)^2 = 9`

or 1 + m² = (1 + m)²

∴ 1 + m² = 1 + 2m + m² 

or 2m = 0

or m = 0

Hence, the slope of line PQ is 0 i.e., the line is parallel to the x-axis.