Question

if the distance from p to the points (5,-4) (7,6) are in the ratio 2:3 then find the locus of p​

Answer 1

Answer:

The locus of P is

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

Step-by-step explanation:

Formula used:

The distance between two points $(x_1,y_1)\:and\:(x_2,y_2)\:is\:d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Let the moving point be P(h,k)

Let the given points be A(5,-4) and B(7,6)

Given:

AP:BP = 2:3

$\frac{AP}{BP}=\frac{2}{3}$

$3\:AP=2\:BP$

$3\:\sqrt{(h-5)^2+(k+4)^2}=2\:\sqrt{(h-7)^2+(k-6)^2}$

squaring on both sides we get

$9\:[(h-5)^2+(k+4)^2]=4\:[(h-7)^2+(k-6)^2]$

$9\:[h^2+25-10h+k^2+16+8k]=4\:[h^2+49-14h+k^2+36-12k]$

$9h^2+225-90h+9k^2+144+72k=4h^2+196-56h+4k^2+144-48k$

$9h^2+225-90h+9k^2+72k=4h^2+196-56h+4k^2-48k$

$5h^2+5k^2-34h+120k+29=0$

$\therefore$ The locus of P is

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