Question

find the equation of lous of P Ratio of distance from p to A (5,-4) & B (7,6) - is 2:3​

Answer 1

$\circledcirc\sf Answer:-$

The equation of locus point P is

$\sf 5x^2+5y^2-34x+120y+29=0$

$\circledcirc \sf Given :-$

The equation of Locus point P the Distance from point A (5, -4) and point B (7,6 ) in ratio 2:3

$\circledcirc \sf To\: find :-$

Equation of locus

$\circledcirc \sf Solution :-$

Let the co-ordinates of point P =(x,y) As, they given Distance between from point A and point B in ratio 2:3

• Let the distance between P and A is PA
• The distance between P and B is PB

According to the Question ,

$\sf \dfrac{PA}{PB} = \dfrac{2}{3}$

We can find distance between them by using Distance formula,

$\sf Distance\: formula = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$\sf PA = \sqrt{(x-5)^2+(y+4)^2}$

$\sf PB = \sqrt{(x-7)^2+(y-6)^2}$

$\sf \dfrac{PA}{PB} = \dfrac{2}{3}$

$\sf \dfrac{\sqrt{(x-5)^2+(y+4)^2}}{\sf\sqrt{(x-7)^2+(y-6)^2}}=\dfrac{2}{3}$

Squaring on both sides

$\bigg(\sf \dfrac{\sqrt{(x-5)^2+(y+4)^2}}{\sf\sqrt{(x-7)^2+(y-6)^2}}\bigg)^2=\bigg(\dfrac{2}{3}\bigg)^2$

$\sf\dfrac{(x-5)^2+(y+4)^2}{(x-7)^2+(y-6)^2} =\dfrac{4}{9}$

Simplifying the numerator and denominator by (a+b)^2 and (a-b)^2

• (a+b)²= a²+2ab+b²
• (a-b)² = a²-2ab+b²

$\sf\dfrac{x^2+25-10x+y^2+16+8y}{x^2+49-14x+y^2+36-12y} =\dfrac{4}{9}$

$\sf\dfrac{x^2+y^2-10x+8y+41}{x^2+y^2+85-14x-12y} =\dfrac{4}{9}$

Do cross multiplication

$\sf{(x^2+y^2-10x+8y+41)}9=({x^2+y^2+85-14x-12y})4$

$\sf{9x^2+9y^2-90x+72y+369}={4x^2+4y^2+340-56x-48y}$

Transpose all terms to L.H.S

$\sf9x^2+9y^2-90x+72y+369-4x^2-4y^2-340+56x+48y=0$

$\sf 5x^2+5y^2-34x+120y+29=0$

So, the equation of locus point P is

$\sf\red{ 5x^2+5y^2-34x+120y+29=0}$

Answer 2

$\large\boxed{\textsf{\textbf{\red{Question\::-}}}}$

• find the equation of lous of P Ratio of distance from p to A (5,-4) & B (7,6) - is 2:3

$\large\boxed{\textsf{\textbf{\red{Answer\::-}}}}$

• Find the equation of locus of P, if the ratio of the distance from P(5,−4) and (7,6) is 2:3

$Let \: P = (x,y)$

$Distance \: from \:  P  \: to  (5,4)= \sqrt{(x - 5 {}^{2} + (y + 4) {}^{2} }$

$Distance \: from \:  P \:  to  \: (7,6)= \sqrt{(x - 7) {}^{2} + (y + 6) {}^{2} }$

$Given \: ratio \: of \: distance =2:3$

$⇒ \frac{ \sqrt{(x - 5) {}^{2} + (y + 4) {}^{2} } }{ \sqrt{(x - 7) {}^{2} + (y - 6) {}^{2} } } = \frac{2}{3}$

squaring on both sides :-

$⇒ \frac{ {x}^{2} + 25 - 10x + {y}^{2} + 16 + 8y }{ {x}^{2} + 49 - 14x + {y}^{2} + 36 - 12y} = \frac{4}{9}$

$⇒9x {}^{2} + 255 - 90x + 9y {}^{2} + 144 + 72y =$

$4x {}^{2} + 196 - 56x + 4y {}^{2} + 144 - 48y$

$⇒5 {x}^{2} + 5y {}^{2} - .34x + 120y + 31 = 0$