Question

Find the equation of the straight line which passes through the point P (2, 6) and cuts the coordinate axes at the point A and B respectively so that \[\frac{AP}{BP} = \frac{2}{3}\] .

Answer 1

The equation of the line with intercepts a and b is \[\frac{x}{a} + \frac{y}{b} = 1\]

Since, the line meets the coordinate axes at A and B, the coordinates of A and B are A (a, 0) and B(0, b).
Given:

\[AP : BP = 2 : 3\]

Here, 

\[P \equiv \left( 2, 6 \right)\]

\[\therefore 2 = \frac{2 \times 0 + 3 \times a}{2 + 3}, 6 = \frac{2 \times b + 3 \times 0}{2 + 3}\]

\[ \Rightarrow 3a = 10, 2b = 30\]

\[ \Rightarrow a = \frac{10}{3}, b = 15\]

Thus, the equation of the line is

\[\frac{x}{\frac{10}{3}} + \frac{y}{15} = 1\]

\[ \Rightarrow \frac{3x}{10} + \frac{y}{15} = 1\]

\[ \Rightarrow 9x + 2y = 30\]