Question

The point on the x-axis which is equidistant from P(-2, 9) and Q (2,-5) is : (a) (0,7) (b) (-7,0) (c) (7,0) (d) (7, -7)​

Answer 1

Answer:

• Option (b) is correct

Step-by-step explanation:

In this question, we are asked to find a point on x-axis which is equidistant from the points P(-2, 9) and Q(2, -5). Inorder to solve this question, we will use distance formula and the concept of general coordinate on x axis.

Any point on x axis is given by (x, 0).

Let's assume that the required coordinate be K(x, 0).

We are aware about the distance formula :

• $\sf Distance\: formula= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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Now,

$\small{ \sf\implies Distance \: of \: PK = Distance \: of \: QK}$

$\small{ \implies\sqrt{(9 - 0)^{2} + ( - 2 - x)^{2} } = \sqrt{( - 5 - 0)^{2} + (2 - x)^{2} } }$

$\small{ \implies\sqrt{81 + ( - 2 - x)^{2} } = \sqrt{25+ (2 - x)^{2} } }$

$\small{ \implies81 + ( - 2 - x)^{2} = 25+ (2 - x)^{2} }$

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Now expanding using algebraic identity :-

• (A + B)² = A² + B² + 2AB
• (A - B)² = A² + B² - 2AB

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$\small{ \implies81 + ( - 2)^{2} + {x}^{2} -2( - 2)( x) = 25+ {2}^{2} + {x}^{2} - 4x }$

$\small{ \implies81 + 4 + {x}^{2} + 4x= 25+ 4 + {x}^{2} - 4x }$

$\small{ \implies85 + 4x= 29 - 4x }$

$\small{ \implies 4x + 4x= 29 - 85 }$

$\small{ \implies 8x= - 56}$

$\small{ \implies x= -7}$

Hence the required coordinates are (-7, 0).

Option (b) is correct