Question
Find the point on the x-axis equidistant from the points $(5,4)$ and $(-2,3).$
✓
Answer
Let the point on x-axis be $P (x,0)$
Given ,
$PA = PB$
$PA^2 = PB^2$
$(x - 5)^2 + (0 - 4)^2 = (x + 2)^2 + (0 - 3)^2$
$x^2 + 25 - 10x + 16 = x^2 + 4 + 4x + 9$
$\Rightarrow - 14 x + 28 = 0$
$\Rightarrow 14 x = 28$
$\Rightarrow x = 2$
$\therefore $ The point on x-axis is $(2 , 0)$
Given ,
$PA = PB$
$PA^2 = PB^2$
$(x - 5)^2 + (0 - 4)^2 = (x + 2)^2 + (0 - 3)^2$
$x^2 + 25 - 10x + 16 = x^2 + 4 + 4x + 9$
$\Rightarrow - 14 x + 28 = 0$
$\Rightarrow 14 x = 28$
$\Rightarrow x = 2$
$\therefore $ The point on x-axis is $(2 , 0)$
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