Question

find the point on x-axis which is equidistant from the point (2,-5)and(-2,9).Also find the area of the isosceles triangle so formed

Answer 1

  Let the point of x-axis be P(x, 0)
Given A(2, -5) and B(-2, 9) are equidistant from P
That is PA = PB
Hence PA2 = PB2  → (1)
Distance between two points is √[(x2 - x1)2 + (y2 - y1)2]
PA = √[(2 - x)2 + (-5 - 0)2]
PA2 = 4 - 4x +x2 + 25 = x2 - 4x + 29
Similarly, PB2 = x2 + 4x + 85
Equation (1) becomes
x2 - 4x + 29 = x2 + 4x + 85
- 8x = 56
x = -7
Hence the point on x-axis is (-7, 0)

Answer 2

Answer:

Step-by-step explanation: given that AP=BP

Then distance AP is also equal to distance BP

By applying distance formula we get x^2- 4x + 29 = x^2 +4x +85

X^2 will cancle out each other

Then,

We get

-4x +29 = 4x + 85

-8x = 56

X= -7