Find the equation of straight line which passes through the point P(2,6) and cuts the coordinate axis at the point A and B respectively so that AP:BP=2:3.

Explanation

As,` AP:BP=2:3`.
By section formula,

`(2,6)=(\frac{3x+0}{5},\frac{2y+0}{5})`

`=(\frac{3x}{5},\frac{2y}{5})`

`\implies x=\frac{10}{3}` and `y=15`

Thus, required line passes through the points `(\frac{10}{3},0)` and `(0,15)`

Hence, the required equation is, `y-0=\frac{15-0}{0-\frac{10}{3}}(x-\frac{10}{3})`

`\implies y=\frac{-9}{2}(\frac{3x-10}{3})`

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