Question
A point $P$ divides the line segment joining the points $A(3, -5)$ and $B(-4, 8)$ such that $\frac{\text{AP}}{\text{PB}}=\frac{\text{k}}{1}.$ If $P$ lies on the line $x + y = 0,$ then find the value of $k.$
✓
Answer
It is given that $\frac{\text{AP}}{\text{PB}}=\frac{\text{k}}{1}$
So, $P$ divides the line segment joining the points $A(3, -5)$ and $B(-4, 8)$ in the ratio $k : 1$
Using the section formula, we get
Coordinates of $\text{P}=\Big(\frac{-4\text{k}+3}{\text{k}+1},\frac{8\text{k}-5}{\text{k}+1}\Big)$
Since $P$ lies on the line $x + y = 0,$ so
$\frac{-4\text{k}+3}{\text{k}+1}+\frac{8\text{k}-5}{\text{k}+1}=0$
$\Rightarrow\ \frac{-4\text{k}+3+8\text{k}-5}{\text{k}+1}=0$
$\Rightarrow\ 4\text{k}-2=0$
$\Rightarrow\ \text{k}=\frac{1}{2}$
Hence, the value of $k$ is $\frac{1}{2}.$
So, $P$ divides the line segment joining the points $A(3, -5)$ and $B(-4, 8)$ in the ratio $k : 1$
Using the section formula, we get
Coordinates of $\text{P}=\Big(\frac{-4\text{k}+3}{\text{k}+1},\frac{8\text{k}-5}{\text{k}+1}\Big)$
Since $P$ lies on the line $x + y = 0,$ so
$\frac{-4\text{k}+3}{\text{k}+1}+\frac{8\text{k}-5}{\text{k}+1}=0$
$\Rightarrow\ \frac{-4\text{k}+3+8\text{k}-5}{\text{k}+1}=0$
$\Rightarrow\ 4\text{k}-2=0$
$\Rightarrow\ \text{k}=\frac{1}{2}$
Hence, the value of $k$ is $\frac{1}{2}.$
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