Solve this

Solution:

Given points are A(a,0),B(0,b) and P(x,y)

For three points to be collinear, the slope of all pairs must be equal, that is the slope of AB = slope of BP = slope of PA.

slope $=\left(\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\right)$

Slope of $A B=\left(\frac{b-0}{0-a}\right)=\frac{b}{-a}$

Slope of BP $=\left(\frac{y-b}{x-0}\right)=\frac{y-b}{x}$

Slope of PA $=\left(\frac{y-0}{x-a}\right)=\frac{y}{x-a}$

Now Slope of AB = BP = PA

$\frac{b}{-a}=\frac{y-b}{x}=\frac{y}{x-a}$

Using the first two equality

$\Rightarrow \frac{\mathrm{b}}{-\mathrm{a}}=\frac{\mathrm{y}-\mathrm{b}}{\mathrm{x}}$

$\Rightarrow \mathrm{bx}=-\mathrm{a}(\mathrm{y}-\mathrm{b})$

$\Rightarrow \mathrm{bx}=-\mathrm{ay}+\mathrm{ab}$

Dividing the equation by “ab”, We get

$\frac{\mathrm{x}}{\mathrm{a}}=-\frac{\mathrm{y}}{\mathrm{b}}+1$

$\Rightarrow \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$

Hence proved.