Class 10

Math

All topics

Coordinate Geometry

If the point $P(2,2)$ is equidistat from the points $A(−2,k)$ and $B(−2k,−3)$, find k. Also, find the length of AP.

Given: Point $P(2,2)$ is equidistant from the two points $A(−2,k)$ and $B(−2k,−3)$

$PA=PB$ or $PA_{2}=PB_{2}$

$(2+2)_{2}+(2−k)_{2}=(2+2k)_{2}(2+3)_{2}$

$4_{2}+4−4k+k_{2}=4+8k+4k_{2}+5_{2}$

$16+4−4k+k_{2}=4+8k+4k_{2}+25$

$4k_{2}+8k+29−20+4k−k_{2}=0$

$3k_{2}+12k+9=0$

$k_{2}+4k+3=0$

$k_{2}+k+3k+3=0$

$k(k+1)+3(k+1)=0$

$(k+1)(k+3)=0$

thus, $k=−1$ or $k=−3$

If $k=−1$

$AP_{2}=20−4k+k_{2}$

$=20+4+1$

$=25$

AP$=5$ units

If $k=−3$

$AP_{2}=20−4k+k_{2}$

$=20+12+9$

$=41$

$AP=41 $ units.