Question

Find the equation of locus of a point P, if the line segment joining (-1, 2) and (3, -2) subtends a right angle at P.​

Answer 1

Answer:

your solution is as follows

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and dude pls don't give such easy ones

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Step-by-step explanation:

$so \: let \: here \\ P = (x,y) \\ \\ A = (x1,y1) = ( - 1,2) \\ B = (x2,y2) = (3, - 2) \\ \\ so \: now \: since \: here \\ ∠ APB = 90 \\ \\ by \: pythagoras \: theorem \\ we \: get \\ \\ AP {}^{2} + PB {}^{2} = AB {}^{2} \\ \\ (x - x1) {}^{2} + (y - y1) {}^{2} + \\ (x - x2) {}^{2} + (y - y2) {}^{2} = (3 + 1) {}^{2} + ( - 2 - 2) {}^{2} \\ \\ (x + 1) {}^{2} + (y - 2) {}^{2} + (x - 3) {}^{2} + \\ (y + 3) {}^{2} = (4) {}^{2} + ( - 4) {}^{2} \\ \\ x {}^{2} + 1 + 2x + y {}^{2} + 4 - 4y+ \\ x {}^{2} + 9 - 6x + y {}^{2} + 9 + 6y = 16 + 16 \\ \\ 2x {}^{2} + 2y {}^{2} - 4x + 2y + 23 = 32 \\ \\ 2x {}^{2} + 2y { }^{2} - 4x + 2y - 9 = 0 \\ \\ or \\ \\ x {}^{2} + y { }^{2} - 2x + y - \frac{9}{2} = 0$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

• The line segment joining (-1, 2) and (3, -2) subtends a right angle at P.

Let assume that

• Coordinates of Locus of point P be (x, y).

• Coordinates of A be ( - 1, 2 )

• Coordinates of B be ( 3, - 2 )

We know,

Slope of a line joining two points (a, b) and (c, d) is

$\red{\rm :\longmapsto\:\boxed{\tt{ Slope \: of \: line \: = \: \frac{d - b}{c - a} \: }}}$

So,

Slope of line segment AP joining the points A(- 1, 2) and P, (x, y) is

$\rm :\longmapsto\:Slope \: of \: AP \: = \: \dfrac{y - 2}{x + 1} - - - (1)$

Again

Slope of line segment BP joining the points B (3, - 2) and P (x, y) is

$\rm :\longmapsto\:Slope \: of \:BP \: = \: \dfrac{y + 2}{x - 3} - - - (2)$

Now, we know that

Two lines having slope m and M are perpendicular iff Mm = - 1

It is given that AB subtends right angle at P

$\rm \implies\:AP \: \perp \: BP$

$\rm \implies\:Slope \: of \: AP \: \times \: Slope \: of \: BP = - 1$

$\rm \implies\:\dfrac{y - 2}{x + 1} \times \dfrac{y + 2}{x - 3} = - 1$

$\rm :\longmapsto \: {y}^{2} - 4 = - (x + 1)(x - 3)$

$\rm :\longmapsto \: {y}^{2} - 4 = - ( {x}^{2} - 3x + x - 3)$

$\rm :\longmapsto \: {y}^{2} - 4 = - ( {x}^{2} - 2x - 3)$

$\rm :\longmapsto \: {y}^{2} - 4 = - {x}^{2} + 2x + 3$

$\rm :\longmapsto \: {x}^{2} + {y}^{2} + 2x - 7 = 0$

So, Locus of point P is a circle.

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