Question

If the point A(2,-4) is equidistant from P(3,8) & Q(-10,y),find the value of y.Also find PQ.​

Answer 1

Hope it is right answer

Answer 2

$\huge{ \underline{ \boxed{ \sf{ \red{Detailed \: AnsweR}}}}}$

Given : The point A(2, -4) is equidistant from P(3,8) & Q(-10,y)

To Find : Find the value of y, Find PQ.

⟹ AP = AQ ( given that they are equidistant )

Let us find their distances using distance formula,

$\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}} = \sqrt{(x2 - x1)^{2} + (y2 - y1)^{2} }$

In AP,

$\sf{x_{1} = 2}$

$\sf{y_{1} = - 4}$

$\sf{x_{2} = 3}$

$\sf{y_{2} = 8}$

In AQ,

$\sf{x_{1} = 2}$

$\sf{x_{2} = - 10}$

$\sf{y_{1} = - 4}$

$\sf{y_{2} = y}$

Substituting the values,

$\sqrt{(3 - 2) ^{2} + (8 + 4} ) ^{2} = \sqrt{( - 10 - 2)^{2} + (y + 4) ^{2} }$

$\sqrt{(1 + 144)} = \sqrt{( - 144) + (y ^{2} - 8y + 16) }$

Squaring on both sides,

$( { \sqrt{(1 + 144)} } \: \: )^{2} = ( { \sqrt{( - 144) + ( {y}^{2} + 8y + 16 } }) ^{2} \: )$

Now square and root gets cancelled,

1 + 144 = - 144 + y² + 8y + 16144

y² + 8y + 16 - 1

= 0y² + 3y + 5y + 15

= 0y ( y + 3 ) + 5 ( y + 3 )

( y + 3 ) ( y + 5 )

$\boxed{\sf{\green{y = -3}}}$$\boxed{\sf{\blue{y = - 5}}}$

These are the values of y.

_________________________

Now we have to find the value of PQ using distance formula.

We got 2 values of y, so we have to check the distance twice.

When the value of y = -3,

P(3,8) & Q(-10,-3)$\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$\sqrt{(-10-3)^2+(-3-8)^2}$$\sqrt{169+121}$

PQ = $\sqrt{290}$

When the value of y = -5,P(3,8) & Q(-10,-5)

$\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$\sqrt{(-10-3)^2+(-5-8)^2}$

$\sqrt{169+169}$PQ = $\sqrt{338}$

______________________