Question

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

Answer 1

Required explaination :

If (x,y) is equidistant from (7,1) and (3,5) then the length of line segment joining these points will also be equal.

Let us assume the coordinates:

• (x,y)=A

• (7,1)=B

• (3,5)=C

Now, the question states that the distance between AB and AC is equal. We know that distance between two coordinates can be find by applying distance Formula, so we can write this statement as::

Distance AB=distance AC---(1)

We have distance formula::

$\leadsto\sf Distance\: formula \bf =\sqrt{(X_2-X_1)^2+(Y_2-Y_1)^2}}$

Here X₁, X₂, Y₁, Y₂ will vary for different coordinates.

Assume coordinates:

• Coordinates of A be X₁ and Y₁

• Coordinates of B be X₂ and Y₂

• Coordinates of C be X₃ and Y₃

Now we can substitute these values for equation (1)

$\dashrightarrow\bf\sqrt{ (X_2-X_1)^2+(Y_2-Y_1)^2}=\sqrt{(X_3-X_1)^2+(Y_3-Y_1)^2}$

By substituting values of coordinates we get::

$\Longrightarrow\sf\sqrt{ (7-X)^2+(1-Y)^2}=\sqrt{(3-X)^2+(5-Y)^2}$

Now applying algeberic identity:

• (A-B)²=A²+B²-2AB

$\Longrightarrow\sf\sqrt{ [7^2+X^2-2(7)(X)]+[1^2+Y^2-2(1)(Y)]}=\sqrt{[3^2+X^2-2(3)(X)]+[(5^2+Y^2-2(5)(Y)]}$

Now solving it::

$\Longrightarrow\sf\sqrt{ [49+X^2-14X]+[1+Y^2-2Y]}=\sqrt{[9+X^2-6X]+[25+Y^2-10Y]}$

$\Longrightarrow\sf\sqrt{ 49+X^2-14X+1+Y^2-2Y}=\sqrt{9+X^2-6X+25+Y^2-10Y}$

$\Longrightarrow\sf\sqrt{ 50+X^2-14X+Y^2-2Y}=\sqrt{34+X^2-6X+Y^2-10Y}$

By squaring both sides square root will be removed from both sides::

$\Longrightarrow\sf 50+{\bf{X^2}}-14X+{\bf{Y^2}}-2Y=34+{\bf{X^2}}-6X+{\bf{Y^2}}-10Y$

X² and Y² will be cancelled out from both sides, we get::

$\Longrightarrow\sf 50-14X-2Y=34-6X-10Y$

Now solving it by transporting of LHS and RHS::

$\Longrightarrow\sf 50-34=14X-6X-10Y+2Y$

$\Longrightarrow\sf 16=8X-8Y$

$\Longrightarrow\sf {\bf{8}}(2)={\bf{8}}(X-Y)$

Cancelling 8 from both sides::

$\boxed{\Longrightarrow\bf 2=X-Y}$

This is the required relation.