Find relation ____ between x and y such that the point (x,y) is equidistant from the point (7,1) and (3,5).

Question

Find relation ____ between x and y such that the point (x,y)  is equidistant from the point (7,1) and (3,5).

Accepted Answer

It is given that the distance of point (x,y) from the point (7,1) and the distance of the point (x,y) from the point(3,5) are equal.
We will first find the distance between the points (x,y) and (7,1)
by applying the distance formula.
We know the distance formula is

\sqrt{[{(x 2 - x 1)}^{2} + {(y 2 - y 1)}^{2}] .}

To find the distance between (x,y) and (7,1) , we will replace x1 by x, y1 by y, x2 by 7 and y2 by 1.
Distance of the point (x,y) from (7,1)

= \sqrt{[{(7 - x)}^{2} + {(1 - y)}^{2}]}

Applying exponents on the bases, we get
⇒ Distance of the point (x,y) from (7,1)

= \sqrt{[49 + x^{2} - 2 \times 7 \times x + 1 + y^{2} - 2 \times 1 \times y]}

On simplifying the equation further, we get
⇒ Distance of the point (x,y) from (7,1)

= \sqrt{[x^{2} + y^{2} - 2 y - 14 x + 50]}

We will apply the same method for finding the distance between (x,y) and (3,5).
We have to find the distance between the points (x,y) and (3,5). For that, we will replace

x 1 by x, y 1 by y, x 2 by 3 and y 2 by 5 .

Distance of the point (x,y) from (3,5)

= \sqrt{[{(3 - x)}^{2} + {(5 - y)}^{2}]}

Applying exponents on the bases, we get
⇒Distance of the point (x,y) from (3,5)

= \sqrt{[9 + x^{2} - 2 \times 3 \times x + 25 + y^{2} - 2 \times 5 \times y]}

On simplifying the equation further, we get
⇒ Distance of the point (x,y) from (3,5) =

\sqrt{[x^{2} + y^{2} - 10 y - 6 x + 34]}

Now, we will equate the distance of point (x,y) from the point (7,1) with the distance of the point (x,y) from the point(3,5). Thus,

\sqrt{[x^{2} + y^{2} - 2 y - 14 x + 50]}

=

\sqrt{[x^{2} + y^{2} - 10 y - 6 x + 34]}

\Rightarrow x^{2} + y^{2} - 2 y - 14 x + 50 = x^{2} + y^{2} - 10 y - 6 x + 34

Taking all the terms to the left hand side, we get

\Rightarrow x^{2} + y^{2} - 2 y - 14 x + 50 - (x^{2} + y^{2} - 10 y - 6 x + 34) = 0

Now, we will open the parenthesis and multiply the negative sign to each term. Therefore, we get

\Rightarrow x^{2} + y^{2} - 2 y - 14 x + 50 - x^{2} - y^{2} + 10 y + 6 x - 34 = 0

On further simplifying the equation, we get
⇒8y−8x+16=0
Taking 8 common from the equation, we get
⇒y−x+2 = 0
This is the required relation between x and y.

Find relation ____ between x and y such that the point (x,y) is equidistant from the point (7,1) and (3,5).

Best Courses for You

Detailed Solution

courses

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

download the app

Download the App