Question

The line segment joining P(5, –2) and Q(9, 6) is divided in the ratio 3 : 1 by a point A on it. Find the equation of a line through the point A parallel to the line \(x\) – 3y + 4 = 0.

Answer 1

Co-ordinates of A are \(\bigg(\frac{3\times9+1\times5}{3+1},\frac{3\times6+1\times-2}{3+1}\bigg)\) = \(\bigg(\frac{32}{4},\frac{16}{4}\bigg)\), i.e. (8, 4)

Now, \(x\) – 3y + 4 = 0 ⇒ –3y = –\(x\) – 4 ⇒ y = \(\frac{x}{3}+\frac{4}{3}\)

∴ Slope of given line = \(\frac{1}{3}\)

⇒ Slope of required line = \(\frac{1}{3}\)      (Since lines are parallel)

∴ Equation of line through (8, 4) with slope = \(\frac{1}{3}\) is

(y - 4) = \(\frac{1}{3}\) (\(x\) - 8)               [Using, y – y₁ = m (x – x₁)]

⇒ 3y – 12 = \(x\) – 8 ⇒ 3y – \(x\) = 4.