A straight line meets the $X$ and $Y$ axes at the points $A, B$ respectively. If $A B=6$ units, then the locus of the point $P$ which divides the line segment $A B$ such that $A P: P B=2: 1$, is

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Aicharya · Accepted Answer

Let $A(a, 0)$ and $B(0, b)$ and point $P(h, k)$ divides the line $A B$. Now, $\begin{aligned} & P(h, k)=\left(\frac{a \times 1+0}{1+2}, \frac{0 \times 2 b}{1+2}\right) \ & \Rightarrow \quad P(h, k)=\left(\frac{a}{3}, \frac{2 b}{3}\right) \ & \therefore \quad h=\frac{a}{3} \text { and } k=\frac{2 b}{3} \ & b=\frac{3 k}{2} \end{aligned}$ $\Rightarrow \quad a=3 h$ and Also, given that $A B=6$ $\begin{aligned} & \sqrt{(a-0)^2+(0-b)^2}=6 \ & \Rightarrow \quad a^2+b^2=36 \ & \Rightarrow \quad 9 h^2+\frac{9 k^2}{4}=36 \ & \Rightarrow \quad h^2+\frac{k^2}{4}=4 \ & \Rightarrow \quad 4 h^2+k^2=16 \end{aligned}$ $\therefore$ Required locus of $P$ is $4 x^2+y^2=16$.

A straight line meets the \(X\) and \(Y\) axes at the points \(A, B\) respectively. If \(A B=6\) units, then…

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