find the equation of locus of p, if the line segment Joining [2, 3] and [-1, 5] subtends a right angle at P.
Answers
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
\begin{gathered}\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}\end{gathered}(x−2)2+(y−3)2=(x+1)2+(y−5)2+(2+1)2+(3−5)2(x−2)2+(y−3)2=(x+1)2+(y−5)2+9+4Squaring both sidesx2−4x+4+y2−6y+9=x2+2x+1+y2−10y+25+13+2×(x+1)2+(y−5)2×13−4x−2x−6y+10y+13−39=2×(x+1)2+(y−5)2×13−6x+4y−26=2×(x+1)2+(y−5)2×13−3x+2y−13=(x+1)2+(y−5)2×13Squaring both sides
\begin{gathered}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\end{gathered}9x2+4y2+169−12xy−52y+78x=13×(x2+2x+1+y2−10y+25)9x2+4y2+169−12xy−52y+78x=13x2+26x+13y2−130y+33813x2−9x2+13y2−4y2+26x−78x+12xy−130y+52y=04x2+9y2−52x−78y+12xy=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=04x2+9y2−52x−78y+12xy=0
Answer:
Let P = (x, y) and A (2, 3), B (-1, 5) be the given points.
Condition is = APB = 90°
➸ PA2 + PB2 = AB2
➸ ( x -2) 2 + (Y-3) 2 + (X+1) 2 + (Y-5) 2 = ( y -5 ) 2 = (-1 -2) 2 + (5-3) 2+ (5 - 3) 2
➸ x2 - 4x + 4 + 4 + y2 - 6y + 9 + x2 + 2x + 1 + y2 - 10y + 25 = 9 + 4
➸2×2 + 2y2 - 2x - 16y + 26 = 0
➸ The local of P is