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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