# solve fast please. Q.8. Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5). Q.9. The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from the points Q (2, - 5) and R (- 3, 6), then find the coordinates of P. Hint: The point P is of the form (2k, k).

Dear Student,

The answers for the questions are as follows:

Using the distance formula to find the distance between two points $P\left({x}_{1},{y}_{1}\right)$ and $Q\left({x}_{2},{y}_{2}\right)$ i.e.
Distance = $\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
Since point (x,y) is equidistant from the points (7,1) and (3,5), hence the distance of the point (x,y) will be same from these two points.

Thus,

Hence, the relation between x and y is;

$x-y=2$

Since, in the question it is given that the x coordinate of the point P is twice the y coordinate.

Hence, the coordinates of the point P will be $\left(2k,k\right)$, where k is the constant.

Now, Using the distance formula to find the distance between two points $P\left({x}_{1},{y}_{1}\right)$ and $Q\left({x}_{2},{y}_{2}\right)$ i.e.
Distance = $\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

Since point $P\left(2k,k\right)$ is equidistant from the points Q(2,-5) and R(-3,6), hence the distance of the point $P\left(2k,k\right)$will be same from these two points.

Thus,

Thus, the coordinates of point $P\left(2k,k\right)$ will be $P\left(16,8\right)$