Question

Find the coordinates of the points which divide, internally and externally, the line joining the point $(a+b,a-b)$ to the point $(a-b,a+b)$ in the ratio $a:b$.

Answer 1

Section formula :
Any point let say (x , y) divides the line joining points $(x_1,y_1)$ & $(x_2,y_2)$ in the ratio $m:n$, then co-ordinates were given by the formula
$x=\frac{x_1\times n+x_2\times m}{m+n}$
$y=\frac{y_1\times n+y_2\times m}{m+n}$
(i) Divides the line internally, take $m$ & $n$ positive
(ii) Divides the line externally, take any one of them negative
Given points are $P(a+b,a-b)$ and $Q(a-b,a+b)$ and ratio $a:b$
(i) Let $R(x^{\prime },y^{\prime })$ divides the line $PQ$ internally
$x^{\prime }=\frac{(a+b)\times b+(a-b)\times a}{a+b}$
$\Rightarrow x^{\prime }=\frac{a^2+b^2}{a+b}$
$y^{\prime }=\frac{(a-b)\times b+(a+b)\times a}{a+b}$
$\Rightarrow y^{\prime }=\frac{a^2-b^2}{a+b}$
So, point $R(\frac{a^2+b^2}{a+b},\frac{a^2+2ab-b^2}{a+b})$ divides the line $PQ$ internally
(ii) Let $S(x^{\prime \prime },y^{\prime \prime })$ divides the line $PQ$ externally
Let us take $n$ negative, then ratio is $a:-b$
$x^{\prime \prime }=\frac{(a+b)\times (-b)+(a-b)\times a}{a-b}$
$\Rightarrow x^{\prime \prime }=\frac{a^2-2ab-b^2}{a-b}$
$y^{\prime \prime }=\frac{(a-b)\times (-b)+(a+b)\times a}{a-b}$
$\Rightarrow y^{\prime \prime }=\frac{a^2+b^2}{a-b}$
So, point $S(\frac{a^2-2ab-b^2}{a-b},\frac{a^2+b^2}{a-b})$ divides the line $PQ$ externally.