Question

Find the coordinates of point on x-axis which is at equal distance from P and Q.​

Answer 1

Answer:

Let the other two point be A(−2,5) and B(2,−3)

So by distance formula we have,

Distance between two points =

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

Then,

PA=PB

PA

2

=PB

2

⇒(a+2)

2

+(0−5)

2

=(a−2)

2

+(0+3)

2

⇒a

2

+4+4a+25=a

2

+4−4a+9

⇒4a+29=−4a+13

⇒ 8a=−16⇒a=−2

∴ The required point isP(−2,0)

Answer 2

The coordinates of the point on the x-axis which is at equal distance from the points P(-2, -5) and Q(2, 3) is ( - 2,0)

Given : The point on the x-axis is at equal distance from the points P(-2, -5) and Q(2, 3) [ Correction in the question ]

To find : The coordinates of the point

Solution :

Step 1 of 3 :

Consider the required point

Let the point on the x-axis is at equal distance from the points P(-2, -5) and Q(2, 3) is (x, 0)

Step 2 of 3 :

Use the condition

By the given condition

$\sf \sqrt{ {(x + 2)}^{2} + {(0 + 5)}^{2} } = \sqrt{ {(x - 2)}^{2} + {(0 - 3)}^{2} }$

Step 3 of 3 :

Find the coordinates of the point

$\sf \sqrt{ {(x + 2)}^{2} + {(0 + 5)}^{2} } = \sqrt{ {(x - 2)}^{2} + {(0 - 3)}^{2} }$

$\sf \implies {(x + 2)}^{2} + {(0 + 5)}^{2} = {(x - 2)}^{2} + {(0 - 3)}^{2}$

$\sf \implies {x}^{2} + 4x + 4 + 25 = {x}^{2} - 4x + 4 + 9$

$\sf \implies 8x + 29 = 13$

$\sf \implies 8x = - 16$

$\sf \implies x = - 2$

Hence the required coordinates of the point is ( - 2,0)

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