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:

Answer 8:

Using the distance formula to find the distance between two points $P(x_1,y_1)$ and $Q(x_2,y_2)$ 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,

$$\begin{gathered} \sqrt{{\left(x-7\right)}^2+{\left(y-1\right)}^2} = \sqrt{{\left(x-3\right)}^2+{\left(y-5\right)}^2} \\ Squaring both the sides, we get, \\ {\left(x-7\right)}^2+{\left(y-1\right)}^2 = {\left(x-3\right)}^2+{\left(y-5\right)}^2 \\ or, \\ {x}^2+7^2-2\left(x\right)\left(7\right)+y^2+1^2-2\left(y\right)\left(1\right)=x^2+3^2-2\left(x\right)\left(3\right)+y^2+5^2-2\left(y\right)\left(5\right) \left\{u\sin g (a-b{)}^2=a^2+b^2-2ab\right\} \\ or, \\ 49-14x+1-2y=9-6x+25-10y \left\{Cancelling the similar terms from both the sides\right\} \\ or, \\ 50-14x-2y=34-6x-10y \\ or, \\ 50-34=14x-6x+2y-10y \\ or, \\ 16=8x-8y \\ or \\ 2=x-y \end{gathered}$$

Hence, the relation between x and y is;

$x-y=2$


Answer 9:

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 $(2k,k)$, where k is the constant.

Now, Using the distance formula to find the distance between two points $P(x_1,y_1)$ and $Q(x_2,y_2)$ i.e.
Distance = $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

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

Thus,

$$\begin{gathered} \sqrt{{\left(2k-2\right)}^2+{\left(k+5\right)}^2} = \sqrt{{\left(2k+3\right)}^2+{\left(k-6\right)}^2} \\ Squaring both the sides, we get, \\ {\left(2k-2\right)}^2+{\left(k+5\right)}^2 = {\left(2k+3\right)}^2+{\left(k-6\right)}^2 \\ or, \\ 4k^2+2^2-2\left(2k\right)\left(2\right)+k^2+5^2+2\left(k\right)\left(5\right)=4k^2+3^2+2\left(2k\right)\left(3\right)+k^2+6^2-2\left(k\right)\left(6\right) \left\{u\sin g (a-b{)}^2=a^2+b^2-2ab and (a+b{)}^2=a^2+b^2+2ab\right\} \\ or, \\ 4-8k+25+10k=9+12k+36-12k \left\{Cancelling the similar terms from both the sides\right\} \\ or, \\ 29+2k=45 \\ or, \\ 2k=16 \\ or, \\ k=8 \end{gathered}$$

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

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards