Question
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 $x^2+y^2$.
✓
Answer
It is given that the point $C(-1,2)$ divides the line segment joining the points $A(2,5)$ and $B(x, y)$ in the ratio $3: 4$ internally.
Using the section formula, we get
$(-1,2)=\left(\frac{3 \times x+4 \times 2}{3+4}, \frac{3 \times y+4 \times 5}{3+4}\right)$
$\Rightarrow(-1,2)=\left(\frac{3 x+8}{7}, \frac{3 y+20}{7}\right)$
$\Rightarrow \frac{3 x+8}{7}=-1 \text { and } \frac{3 y+20}{7}=2$
$\Rightarrow 3 x+8=-7 \text { and } 3 y+20=14$
$\Rightarrow 3 x=-15 \text { and } 3 y=-6$
$\Rightarrow x=-5 \text { and } y=-2$
$\therefore x^2+y^2=25+4=29$
Hence, the value of $x^2+y^2$ is $29$ .
Using the section formula, we get
$(-1,2)=\left(\frac{3 \times x+4 \times 2}{3+4}, \frac{3 \times y+4 \times 5}{3+4}\right)$
$\Rightarrow(-1,2)=\left(\frac{3 x+8}{7}, \frac{3 y+20}{7}\right)$
$\Rightarrow \frac{3 x+8}{7}=-1 \text { and } \frac{3 y+20}{7}=2$
$\Rightarrow 3 x+8=-7 \text { and } 3 y+20=14$
$\Rightarrow 3 x=-15 \text { and } 3 y=-6$
$\Rightarrow x=-5 \text { and } y=-2$
$\therefore x^2+y^2=25+4=29$
Hence, the value of $x^2+y^2$ is $29$ .
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