Question

solve this question plzz mate ​

Answer 1

Answer:

B(0,3)

answer for the given problem is given

Answer 2

Step-by-step explanation:

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

ㅤ ㅤ 2. ㅤ ㅤ 1

•__________•__________•

A(2,5).ㅤ ㅤ B(x,y). ㅤ C(-1,2)

Given :

Points are A(2,5),B(x,y) & C(-1,2) are collinear such that AB=2BC

AB=2BC

$\frac{AB}{BC} = \frac{2}{1}$

$\bold{\boxed{∴AB:BC=2:1}}$

Let m1:m2=2:1

Section formula

$x = \frac{m1x2 + m2x1}{m1 + m2}$

$y = \frac{m1y2 + m2y1}{m1 + m2}$

(x1 ,y1)=(2,5)

(X2,y2=(-1,2)

$B(x,y) = \frac{[(2)( - 1) + (1)(2)}{2 + 1}, \frac{(2)(2) + (1)(5)]}{2 + 1}$

$B(x,y) = \frac{ - 2 + 2}{3} , \frac{4 + 5}{3} = ( \frac{0}{3}, \frac{9}{3} )$

Hence ,the Points are B(x,y)=(0,3)