Question

find the co ordinates of the points which divides the line segment joining the poits (a+b,a-b)and (a-b,a+b)in the ratio 3:2 INTERNALLY

Answer 1

Answer:

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

Step-by-step explanation:

Let the points be A(a+b, a-b) & B(a-b, a+b).

Given ratio =3:2

By section formula,

$x = \frac{mx2 + nx1}{m + n}$

$y = \frac{my2 + ny1}{m + n}$

Here m:n=3:2

So on putting the values we get,

$x = \frac{3(a - b) + 2(a + b)}{3 + 2}$

$x = \frac{3a - 3b + 2a + 2b}{5} \\ x = \frac{5a - b}{5}$

And,

$y = \frac{3(a + b) + 2(a - b)}{3 + 2}$

$y = \frac{3a + 3b + 2a - 2b}{5}$

$y = \frac{5a + b}{5}$

Therefore the coordinates of the points are :

$( \frac{5a - b}{5} ) \: and \: ( \frac{5a + b}{5} )$

Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

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)