If the point $C (-1, 2)$ divides internally the line segment joining the points $A (2, 5)$ and $B (x, y)$ in the ratio $3 : 4$, find the value of $x^2 + y^2$.

Given:

Point $C (-1, 2)$ divides internally the line segment joining the points $A (2, 5)$ and $B (x, y)$ in the ratio $3 : 4$.

To do:

We have to find the value of $x^2 + y^2$.

Solution:

Using the section formula, if a point $( x,\ y)$ divides the line joining the points

$( x_1,\ y_1)$ and $( x_2,\ y_2)$ in the ratio $m:n$, then 

$(x,\ y)=( \frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n})$

This implies,

$C(-1,\ 2)=( \frac{3(x)+4(2)}{3+4},\ \frac{3(y)+4(5)}{3+4})$

On comparing, we get,

$-1=\frac{3x+8}{7}$

$\Rightarrow -1(7)=3x+8$

$\Rightarrow 3x=-7-8$

$\Rightarrow 3x=-15$

$\Rightarrow x=\frac{-15}{3}$

$\Rightarrow x=-5$

$2=\frac{3y+20}{7}$

$\Rightarrow 2(7)=3y+20$

$\Rightarrow 3y=14-20$

$\Rightarrow 3y=-6$

$\Rightarrow y=\frac{-6}{3}$

$\Rightarrow y=-2$

Therefore,

$x^2+y^2=(-5)^2+(-2)^2$

$=25+4$

$=29$

The value of $x^2+y^2$ is $29$.

Tutorialspoint

Updated on: 10-Oct-2022

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