Question

find the point on the x axis which is equivalent from (2,-5) & (-2,9)​

Answer 1

Answer:

The point is (-7, 0)

Step-by-step explanation:

We need a point on x axis such that it is equidistant from (2, -5) and (-2, 9)

Let A = (2, -5) = (x1, y1)

B = (-2, 9) = (x2, y2)

Now since the point is on the x axis,

it will definitely be of the form P(x, 0) = (x3, y3)

Now, using distance formula,

AP = BP

√[(x3 - x1)² + (y3 - y1)²] = √[(x3 - x2)² + (y3 - y1)²]

√[(x - 2)² + (0 - (-5))²] = √[(x - (-2))² + (0 - 9)²]

√(x² - 4x + 4) + (25) = √(x² + 4x + 4) + (81)

Now squaring on both sides we get,

x² - 4x + 4 + 25 = x² + 4x + 4 + 81

x² - 4x + 29 = x² + 4x + 85

x² - x² - 4x + 29 = 4x + 85

29 - 4x = 4x + 85

4x + 85 + 4x = 29

8x = 29 - 85

8x = (-56)

x = (-56)/8

x = (-7)

Thus,

P = (x, 0) = (-7, 0)

Hence,

The point is (-7, 0)

Hope it helped and believing you understood it.... All the best