Question

(p, q) and (q, p) internally in the ratio p-q:p+q​

Answer 1

Given that :-

( p, q) and (q , p) divides internally in ratio p- q : p+ q

To find :-

The co-ordinates of the point

Solution:-

If the point ${(x_1,y_1)}$ and ${(x_2,y_2)}$ divides the line segemnt in ratio m:n then the co-ordinates are

Section formula Internal division :-

$\bigg(\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}\bigg)$

${x_1 = p}$ ${x_2 = q}$ ${y_1 = q}$ ${y_2 =p}$

• m = p-q
• n = p+q

Substituing the values :-

$\bigg(\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}\bigg)$

$\bigg(\dfrac{(p-q)(q)+(p+q)(p)}{p-q +p+q}, \dfrac{(p-q)(p)+(p+q)(q)}{p-q+p+q}\bigg)$

$\bigg(\dfrac{pq-q^2+p^2+pq}{2p}, \dfrac{p^2-pq+pq+q^2}{2p}\bigg)$

$\bigg(\dfrac{-q^2+p^2+2pq}{2p}, \dfrac{p^2+q^2}{2p}\bigg)$

$\bigg(\dfrac{p^2+2pq-q^2}{2p}, \dfrac{p^2+q^2}{2p}\bigg)$-Required answer

So, the point $\bigg(\dfrac{p^2+2pq-q^2}{2p}, \dfrac{p^2+q^2}{2p}\bigg)$ divides the linesegemnt (p,q) and (q,p) internally in ratio p-q : p+q

__________________

Know more :-

Distance formula:-

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Centroid formula:-

$\bigg(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\bigg)$

Section formula External division:-

$\bigg(\dfrac{mx_2-nx_1}{m-n}, \dfrac{my_2-ny_1}{m-n}\bigg)$

Mid point formula:-

$\bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2}\bigg)$