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 = ${\displaystyle -\frac{\text{m}-2}{\text{m}+1}}$

Now y = 4 − x

= ${\displaystyle 4+\frac{\text{m}-2}{\text{m}+1}}$

= ${\displaystyle \frac{4\text{m}+4+\text{m}-2}{\text{m}+1}=\frac{5\text{m}+2}{\text{m}+1}}$

Given: PQ = 3 or PQ² = 9

∴ ${\displaystyle {(-\frac{\text{m}-2}{\text{m}+1}+1)}^2+{(\frac{5\text{m}+2}{\text{m}+1}-2)}^2=9}$

or ${\displaystyle {\left(\frac{-\text{m}+2+\text{m}+1}{\text{m}+1}\right)}^2+{\left(\frac{5\text{m}+2-2\text{m}-2}{\text{m}+1}\right)}^2=9}$

or ${\displaystyle \frac{9}{{(\text{m}+1)}^2}+{\left(\frac{3\text{m}}{\text{m}+1}\right)}^2=9}$

or ${\displaystyle \frac{9+9{\text{m}}^2}{{(\text{m}+1)}^2}=9}$

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.