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If $A(5,-4)$ and $B(7,6)$ are points in a plane, then the set of all points $P(x, y)$ in the plane such that $A P: P B=2: 3$ is
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Verified Answer
The correct answer is:
a circle
Given, $A(5,-4)$ and $B(7,6)$ are the points in a plane.
Now,
and
$$
\begin{aligned}
& A P=\sqrt{(x-5)^2+\left[\left(y-(-4)^2\right]\right.} \\
& P B=\sqrt{(x-7)^2+(y-6)^2}
\end{aligned}
$$
According to the question,
$$
A P: P B=2: 3
$$
$$
\Rightarrow \quad \frac{\sqrt{(x-5)^2+(y+4)^2}}{\sqrt{(x-7)^2+(y-6)^2}}=\frac{2}{3}
$$
On squaring both sides, we get
$$
\begin{array}{ccc}
\Rightarrow & \frac{(x-5)^2+(y+4)^2}{(x-7)^2+(y-6)}=\frac{4}{9} \\
\Rightarrow & 9\left(x^2+y^2-10 x+8 y+41\right) \\
& =4\left(x^2+y^2-14 x-12 y+85\right) \\
\therefore & 5 x^2+5 y^2-34 x+120 y+29=0
\end{array}
$$
Which represents the equation of an circle.
Now,
and
$$
\begin{aligned}
& A P=\sqrt{(x-5)^2+\left[\left(y-(-4)^2\right]\right.} \\
& P B=\sqrt{(x-7)^2+(y-6)^2}
\end{aligned}
$$
According to the question,
$$
A P: P B=2: 3
$$
$$
\Rightarrow \quad \frac{\sqrt{(x-5)^2+(y+4)^2}}{\sqrt{(x-7)^2+(y-6)^2}}=\frac{2}{3}
$$
On squaring both sides, we get
$$
\begin{array}{ccc}
\Rightarrow & \frac{(x-5)^2+(y+4)^2}{(x-7)^2+(y-6)}=\frac{4}{9} \\
\Rightarrow & 9\left(x^2+y^2-10 x+8 y+41\right) \\
& =4\left(x^2+y^2-14 x-12 y+85\right) \\
\therefore & 5 x^2+5 y^2-34 x+120 y+29=0
\end{array}
$$
Which represents the equation of an circle.
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