Question

41. IfC(-1,2 ) divides the line segment connecting A(2,5) and B(x,y) in a ratio 3:4, then find the value
of x² + y²​

Answer 1

$\large {\textsf{\textbf{Given:}}}$

Given that point C(-1,2) divides the line segment connecting A(2,5) and B(x,y) in a ratio 3:4.

$\large {\textsf{\textbf{To \: find:}}}$

We have to find the value of x²+y².

$\large {\textsf{\textbf{Solution:}}}$

Using the section formula :

$\boxed{\pmb{\sf{( \cfrac{m_1x_2 + m_2x_1}{m_2+m_1} , \cfrac{m_2y_2+m_2y_1}{m_2+m_1} )}}}$

Here,

• $\sf m_1=3$

• $\sf m_2=4$

• $\sf x_1=2$

• $\sf x_2=x$

• $\sf y_1=5$

• $\sf y_2=y$

$\: \: \sf :\implies(-1,2)=( \cfrac{3x + 4(2)}{3 + 4} , \cfrac{3y + 4(5)}{3+4}$

$\: \: \sf :\implies (-1,2)=( \cfrac{3x + 8}{7} , \cfrac{3y + 20}{7} )$

$\:\:\sf:\implies \cfrac{3x + 8}{7} = - 1~;~ \cfrac{3y + 20}{7} = 2$

Solving for x and y :

$\:\:\sf :\implies 3x+8=-1(7)$

$\: \: \sf : \implies3x + 8 = - 7$

$\: \: \sf:\implies 3x=-7-8$

$\: \: \sf:\implies x= \cfrac{-15}{3}$

$\boxed {\pmb{\sf \therefore \: x=-5}}$

$\: \: \sf : \implies3y + 20 = 2(7)$

$\: \: \sf:\implies 3y+20=14$

$\: \: \sf :\implies 3y=14-20$

$\: \: \sf :\implies y= \cfrac{-6}{3}$

$\boxed{\pmb{\sf \therefore \: y=-2}}$

Now, substituting and finding x²+y² :

$\: \: \sf :\implies x^2+y^2$

$\: \: \sf :\implies ( - 5)^2+(-2)^2$

$\: \: \sf :\implies 25+4$

$\boxed {\pmb{\sf \therefore \: x^2+y^2=29}}$