Question
A point $P$ divides the line segment joining the points $A (3,-5)$ and $B (-4,8)$ such that $\frac{A P}{P B}=\frac{k}{1}$. If $P$ lies on the line $x+y=0$, then find the value of $k$.
✓
Answer
Given points are $A (3,-5)$ and $B (-4,8)$.
$P$ divides $AB$ in the ratio $k : 1$
Using the section formula, we have:
Coordinate of point $P$ are $\left\{\left(\frac{-4 k+3}{k+1}\right)\left(\frac{8 k-5}{k+1}\right)\right\}$
Now it is given, that $P$ lies on the line $x+y=0$
Therefore,
$\frac{-4 k+3}{k+1}+\frac{8 k-5}{k+1}=0$
$\Rightarrow-4 k+3+8 k-5=0$
$\Rightarrow-4 k+3+8 k-5=0$
$\Rightarrow 4 k-2=0$
$\Rightarrow k=\frac{2}{4}$
$\Rightarrow k=\frac{1}{2}$
Thus, the value of $k$ is $1 / 2$.
$P$ divides $AB$ in the ratio $k : 1$
Using the section formula, we have:
Coordinate of point $P$ are $\left\{\left(\frac{-4 k+3}{k+1}\right)\left(\frac{8 k-5}{k+1}\right)\right\}$
Now it is given, that $P$ lies on the line $x+y=0$
Therefore,
$\frac{-4 k+3}{k+1}+\frac{8 k-5}{k+1}=0$
$\Rightarrow-4 k+3+8 k-5=0$
$\Rightarrow-4 k+3+8 k-5=0$
$\Rightarrow 4 k-2=0$
$\Rightarrow k=\frac{2}{4}$
$\Rightarrow k=\frac{1}{2}$
Thus, the value of $k$ is $1 / 2$.
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