Question

Find the direction in which a straight line must be drawn through the point (– 1, 2) so that its be at a distance of 3 units from this point.

Answer 1

Let ‘m’ be the slope of the line passing through (– 1, 2).

So, its equation is

y – 2 = m (x + 1) ⇒ y = mx + m + 2 …(1)

Putting the value of y from (1) in x + y = 4, we have

x + mx + m + 2 = 4

⇒ x = \(\frac{2 − m}{1 + m}\)

(1) ⇒ y = m(\(\frac{2 − m}{1 + m}\)) + m + 2

⇒ y = \(\frac{2m − m^2+m^2+3m+2}{1 + m}\)

⇒ y = \(\frac{5m + 2}{1 + m}\)

∴ The co-ordinates of Q, the point of intersection of the lines is (\(\frac{2 − m}{1 + m}\), \(\frac{2 + 5m}{1 + m}\)).

The point P is (– 1, 2)

∴ 3² = \(\big(\frac{2-m}{1+m}+1\big)^2\) + \(\big(\frac{2+5m}{1+m}-2\big)^2\)

⇒ \(\big(\frac{3}{1+m}\big)^2\) + \(\big(\frac{3m}{1+m}\big)^2\) = 9

⇒ 1 + m² = (1 + m)₂

⇒ 1 + m₂ = 1 + m₂ + 2m

⇒ m = 0

∴ Slope of the line is 0 i.e., it is parallel to x-axis.