A straight line passes through P(2, 1) and cuts the axes in points A, B. If BP : PA = 3 : 1.

Find :

(i) the coordinates of A and B.

(ii) the equation of the line AB.

A lies on x-axis and B lies on y-axis. Let coordinates of A be (x, 0) and B be (0, y) and P(2, 1) divides BA in the ratio 3 : 1.

By section formula,

$\text{x-coordinate } = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] \Rightarrow 2 = \dfrac{3x + 0}{3 + 1} \\[1em] \Rightarrow 2 = \dfrac{3x}{4} \\[1em] \Rightarrow x = \dfrac{8}{3}.

Similarly,

$\text{y-coordinate } = \dfrac{m$1y2 + m2y1}{m1 + m2} \\[1em] \Rightarrow 1 = \dfrac{3 \times 0 + 1 \times y}{3 + 1} \\[1em] \Rightarrow 1 = \dfrac{y}{4} \\[1em] \Rightarrow y = 4.

Hence, coordinates of A are $\Big(\dfrac{8}{3}, 0\Big)$ and of B are (0, 4).

(ii) By two point formula equation of AB will be,

$\Rightarrow y - y$1 = \dfrac{y2 - y1}{x2 - x1}(x - x1) \\[1em] \Rightarrow y - 0 = \dfrac{4 - 0}{0 - \dfrac{8}{3}}(x - \dfrac{8}{3}) \\[1em] \Rightarrow y = \dfrac{4 \times 3}{-8}\Big(\dfrac{3x - 8}{3}\Big) \\[1em] \Rightarrow y = \dfrac{3x - 8}{-2} \\[1em] \Rightarrow -2y = 3x - 8 \\[1em] \Rightarrow 3x + 2y - 8 = 0.

Hence, the equation of the required line is 3x + 2y = 8.