Question

If $R(x,y)$ is a point on the line segment joining the points $P(a,b)$ and $Q(b,a)$, then prove that $x+y=a+b$

Answer 1

Given that, $R(x,y)$ divides $PQ$ in the ratio $k:1$

Then we have,

$R(X,Y)=(\frac{k{x}_1+x_2}{k+1},\frac{k{y}_1+y_2}{k+1})$

Here, $x_1=a,y_1=b,x_2=b,y_2=a$

Then $P\left(x,y\right)=\left(\frac{bk+a}{k+1},\frac{ak+b}{k+1}\right)$

$\Rightarrow x=\frac{bk+a}{k+1}$ and $\text{y =}\left(\frac{ak+b}{k+1}\right)$

$\Rightarrow kx+x=bk+a$ and $\text{yk + y = ak + b}$

$\Rightarrow k\left(x-b\right)=a-x$ $\Rightarrow \text{k}\left(y-a\right)=b-y$

$\Rightarrow k=\frac{a-x}{x-b}$ ---(i) $\Rightarrow \text{k =}\left(\frac{b-y}{y-a}\right)$ ---(ii)

from (i) and (ii)

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

$\Rightarrow ay-a^2-xy+ax=bx-b^2+by-xy$

$\Rightarrow \left(a-b\right)y+\left(a-b\right)x-\left(a^2-b^2\right)=0$

$\Rightarrow \left(a-b\right)[y+x-\left(a+b\right)]=0$

$\Rightarrow x+y-\left(a+b\right)=0$

$\Rightarrow x+y=a+b$

Hence proved