Question

Find equation of locus of p if the line segment joining (2,3)and(-1,5)subtends a right angle at p

Answer 1

Answer:

Let Locus of point P be (x,y).

It is given that  line segment joining (2,3)and(-1,5)subtends a right angle at p.

By Pythagorean theorem

(Hypotenuse)²= (Base)² + (Altitude)²

Using Distance formula

$\sqrt{(x-2)^2+(y-3)^2}=\sqrt{(x+1)^2+(y-5)^2}+\sqrt{(2+1)^2+(3-5)^2}\\\\\sqrt{(x-2)^2+(y-3)^2}=\sqrt{(x+1)^2+(y-5)^2}+\sqrt{9+4}\\\\ \text{Squaring both sides}\\\\x^2-4x+4+y^2-6y+9=x^2+2 x+1+y^2-10 y+25+13 +2 \times \sqrt{(x+1)^2+(y-5)^2} \times \sqrt{13}\\\\ -4 x -2 x -6 y+10 y+13 -3 9=2 \times \sqrt{(x+1)^2+(y-5)^2} \times \sqrt{13}\\\\ -6 x + 4y -26=2 \times \sqrt{(x+1)^2+(y-5)^2} \times \sqrt{13}\\\\ -3 x+2 y -13=\sqrt{(x+1)^2+(y-5)^2} \times \sqrt{13}\\\\ \text{Squaring both sides}$

$9x^2+4 y^2+169-12 xy - 52 y+78 x=13 \times (x^2+2 x+1+y^2-10 y+25)\\\\ 9x^2+4 y^2+169-12 xy - 52 y+78 x=13x^2+26 x +13 y^2-130 y+338\\\\13 x^2 - 9 x^2+13y^2-4y^2+2 6 x- 78 x +12 xy -130 y +52 y=0\\\\4x^2+9 y^2-52 x-78 y+12 x y=0$

So, Locus of point p, having coordinate (x,y) is given by:

                        $4x^2+9 y^2-52 x-78 y+12 x y=0$

Answer 2

Answer:

hope it helps you

Step-by-step explanation:

Let P = (x, y) and A(2, 3), B(–1, 5) be the given points.

Given condition is =>APB = 90°

=> PA2 + PB2 = AB2

=> (x – 2)2 + (y – 3)2 + (x + 1)2 + (y – 5)2 = (–1 – 2)2 + (5 – 3)2

=> x2 – 4x + 4 + y2 – 6y + 9 + x2 + 2x + 1 + y2 – 10y + 25 = 9 + 4

=> 2x2 + 2y2 – 2x – 16y + 26 = 0

=> x2 + y2 – x – 8y + 13 = 0

=> The locus of P is

x2 + y2 – x –8y+13=0