Question

Find a point on x-axis which s equidstant from A(5,4) and B(-2,3)

Answer 1

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Given that the point will be on x - axis, so co ordinate of y - axis will be 0 .



Let  point on x-axis which is equidistant from A( 5 , 4 ) and B( - 2 , 3 ) be k ,



By Distance Formula,


⇒ distance between A and ( k , 0 ) = distance between ( k , 0 ) and B


⇒ $\sqrt{( x_{2} - x_{1} )^{2} + (y_{2} - y_{1} )^{2} } = \sqrt{( x_{2} - x_{3} )^{2} + ( y_{2} - y_{3})^{2}}$


⇒$( k - 5 )^{2} + ( 0 - 4 )^{2} = ( k + 2 )^{2} + ( 0 - 3 )^{2} \\ \\ \\  k^{2} + 25 - 10k + 16 = k^{2} + 4 + 4k + 9 \\ \\ \\ k^{2} - k^{2} + 25 + 16- 4 - 9 - 10k - 4k = 0 \\ \\ \\ 28 - 14k = 0 \\ \\ \\ 14( 2 - k ) = 0 \\ \\ \\ 2 - k = 0 \\ \\ \\ 2 = k$



$\boxed{ \boxed{\underline{ \textbf{Hence the point which will be equidistant from A and B on x - axis is ( k , 0 ) = ( 2 , 0 ).}}}}$

Answer 2

Hey there!

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Q. Find the point on the x - axis which is equidistant from (5, 4) and (-2, 3).

Solution :

Since the point on x - axis have its ordinate = 0.

So, P(x, 0) is any point on the x - axis.

Since P(x, 0) is equidistant from A(5, 4) and B(-2, 3).

PA = PB ⇒ $PA^{2} = PB^{2}$

⇒ $(x - 5)^{2} + (0 - 4)^{2} = (x + 2)^{2} + (0 - 3)^{2}$

⇒  $x^{2}$ + 25 - 10x + 16 = $x^{2}$ + 4 + 4x +9

⇒ $x^{2}$ + 41 - 10x = $x^{2}$ + 13 + 4x

⇒ -10x - 4x = 13 - 41

⇒ - 14x = - 28

⇒ $x = \frac{-28}{-14}$ = 2

∴ The point equidistant from given points on the x - axis is (2, 0).