If two points P(x,y) is equidistant …
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Posted by Arshdeep Kaur 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
|PQ| = |PR
{tex}\begin{aligned} \sqrt { [ x - ( a + b ) ] ^ { 2 } + [ y - ( b - a ) ] ^ { 2 } } = \sqrt { [ x - ( a - b ) ] ^ { 2 } + [ y - ( b + a ) ] ^ { 2 } } \end{aligned}{/tex}
Squaring, we get
[x - (a + b)]2 + [y - (b - a)]2 = [x - (a - b)]2 + [y - (a + b)]2
or, [x - (a + b)]2 - [x - a + b]2 = (y - a - b)2 - (y - b + a)2
or, (x - a - b + x - a + b) ( x - a - b - x + a - b)
= (y - a - b + y - b + a)(y - a - b - y + b - a)
or, (2x - 2a) (- 2b) = (2y - 2b) (- 2a)
or, (x - a)b = (y - b)a
or, bx = ay.
Hence Proved.
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