Question

The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2,-5) and R(-3,6). Find the coordinates of P

Answer 1

ans is (16,8)....,,,,,,,,,,,,, ,

Answer 2

Answer:

(16,8)

Step-by-step explanation:

The x-coordinate of a point P is twice its y-coordinate.

Let the y coordinate of P be a

So, x-coordinate of a point P is 2a

Coordinates of P =(2a,a)

Now If P is equidistant from Q(2,-5) and R(-3,6).

So, PQ = QR

We will use distance formula :

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

To find PQ

$(x_1,y_1)=(2a,a)$

$(x_2,y_2)=(2,-5)$

$PQ=\sqrt{(2-2a)^2+(-5-a)^2}$

To find PR

$(x_1,y_1)=(2a,a)$

$(x_2,y_2)=(-3,6)$

$PR=\sqrt{(-3-2a)^2+(6-a)^2}$

Since  PQ = QR

So, $\sqrt{(2-2a)^2+(-5-a)^2}=\sqrt{(-3-2a)^2+(6-a)^2}$

$(2-2a)^2+(-5-a)^2=(-3-2a)^2+(6-a)^2$

$4+4a^2-8a+25+a^2+10a=9+4a^2+12a+36+a^2-12a$

$4-8a+25+10a=9+36$

$29+2a=45$

$2a=16$

$a=8$

So, Coordinates of P =(2a,a) = (2(8),8)=(16,8)

Hence the coordinates of P is (16,8)