Question

Find the coordinates of the point which divides the line segment joining the points
(a+b, a-b) and (a-b, a+b) in the ratio 3 : 2 internally.​

Answer 1

Solution

Refer to attachment

Let the points if the line segment be A(a + b, a - b) and B(a - b, a + b)

Let P be the point which divides the line segment in the ratio of 3 : 2.

P divides AB internally in the ratio of 3 : 2

The coordinates of P can be found by using Section formula

$\therefore A(a +b ,a - b) \ B(a - b,a + b) \ m_1:m_2 = 3:2$

$P(x,y) = \bigg( \dfrac{m_1x _2 + m_2x_1}{m_1 + m_2} , \dfrac{m_1y _2 + m_2y_1}{m_1 + m_2} \bigg)$

$\implies P(x,y) = \bigg( \dfrac{3(a - b) + 2(a + b)}{3 + 2} , \dfrac{3(a + b) + 2(a - b)}{3 + 2} \bigg)$

$\implies P(x,y) = \bigg( \dfrac{3a - 3b+ 2a + 2b}{5} , \dfrac{3a + 3b + 2a - 2b}{5} \bigg)$

$\implies \boxed{P(x,y) = \bigg( \dfrac{5a - b}{5} , \dfrac{5a + b}{5} \bigg)}$

$\text{ Hence, the coordinates are } \bigg( \dfrac{5a - b}{5} , \dfrac{5a + b}{5} \bigg)$

Answer 2

Step-by-step explanation:

answer become (5a-b/5),(5a+b/5)