Question:
Find the point on the y-axis which is equidistant from the points A(6, 5) and B(− 4, 3).
Solution:
Let P (0, y) be a point on the y-axis. Then as per the question, we have
$A P=B P$
$\Rightarrow \sqrt{(0-6)^{2}+(y-5)^{2}}=\sqrt{(0+4)^{2}+(y-3)^{2}}$
$\Rightarrow \sqrt{(6)^{2}+(y-5)^{2}}=\sqrt{(4)^{2}+(y-3)^{2}}$
$\Rightarrow(6)^{2}+(y-5)^{2}=(4)^{2}+(y-3)^{2}$ (Squaring both sides)
$\Rightarrow 36+y^{2}-10 y+25=16+y^{2}-6 y+9$
$\Rightarrow 4 y=36$
$\Rightarrow y=9$
Hence, the point on the y-axis is (0, 9).
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.