Find the equation of locus of a point in which the line segment $A (1, 2)$ and $B (- 3, 4)$ is $90^{o}$

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Solution

Required locus is a circle with AB as diameter,
$∴$ Centre $= \frac{- 3 + 1}{2}, \frac{4 + 2}{2}$

$= \frac{- 2}{2}, \frac{6}{2}$

$= (- 1, 3)$

$\Rightarrow A B = \sqrt{{(- 3 - 1)}^{2} + {(4 - 2)}^{2}} = \sqrt{16 + 4} = \sqrt{20}$
$∴$ Radius of circle $= \sqrt{5}$
$∴$ Equation $= {(x + 1)}^{2} + {(y - 3)}^{2} = 5.$
Hence, the answer is $5.$

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Find the equation of locus of a point in which the line segment A(1,2) and B(−3,4) is 90o

Required locus is a circle with AB as diameter,∴ Centre =−3+12,4+22 =−22,62 =(−1,3)⇒AB=√(−3−1)2+(4−2)2=√16+4=√20∴ Radius of circle =√5∴ Equation =(x+1)2+(y−3)2=5.Hence, the answer is 5.

Find the equation of locus of a point in which the line segment $A (1, 2)$ and $B (- 3, 4)$ is $90^{o}$