Question

Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).​

Answer 1

Answer:

Let the point on the x-axis be (a,b). But we know that, point on the x-axis always has it's ordinate (y-value) to be zero. Hence (b) = 0.

Therefore the unknown point is (a,0)

It is given that this point equidistant from points (2,-5) and (-2,9). Hence applying distance formula separately we get:

Distance b/w (2,-5) & (a,0) = Distance b/w (-2,9) & (a,0)

⇒ ( a - 2 )² + ( 0 + 5 )² = ( a + 2 )² + ( 0 - 9 )²

⇒ ( a² - 4a + 4 ) + 25 = ( a² + 4a + 4 ) + 81

⇒ a² - 4a + 29 - a² - 4a - 85 = 0

⇒ - 8a - 56 = 0

⇒ 8a = - 56

⇒ a = - 56 / 8

⇒ a = -7

Hence the point from which (2,-5) & (-2,9) are equidistant is (-7,0).

Answer 2

Step-by-step explanation:

Let A(2, -5) and B(-2, 9)

Let C(a , 0) be the point which is equidistant from (2, -5) and (-2, 9)

The formula for finding the distance between two points is:-

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

$\sf AC = \sqrt{(a - 2) ^{2} + \{0 -(-5)\} ^{2} }$

$\sf AC = \sqrt{(a - 2) ^{2} + 5^{2}}$

$\underline{\sf AC = \sqrt{(a - 2) ^{2} + 25}}$

$\sf BC = \sqrt{ \{a - (-2) \} ^{2} + (0 - 9)^{2} }$

$\sf BC = \sqrt{(a + 2) ^{2} + 9 ^{2}}$

$\underline{\sf BC = \sqrt{(a + 2) ^{2} + 81}}$

As, AC and and BC are equidistant from the point C

$\sf\dashrightarrow AC = BC$

$\sf \dashrightarrow \sqrt{(a - 2) ^{2} + 25} = \sqrt{(a + 2) ^{2} + 81}$

$\sf Squaring \: on \: both \: sides$

$\sf \dashrightarrow (a - 2) ^{2} + 25 = (a + 2) ^{2} + 81$

$\sf \dashrightarrow a ^{2} + 4 - 4a + 25 = {a}^{2} + 4 + 4a + 81$

$\sf \dashrightarrow - 4a - 4a = 81 - 25$

$\sf \dashrightarrow - 8a = 56$

$\sf \dashrightarrow a = \dfrac{56}{ - 8}$

$\dashrightarrow \underline{\boxed{\sf a = -7}}$

∴ The point on the x-axis which is equidistant from (2, -5) and (-2, 9) is (-7 , 0)