The value of the coordinate x in the point P(x, 3) s -2
From the question, we have the following coordinate points
Point P = (x, 3)
Point Q = (3, 5)
Point R = (-4, -2)
The point P is equidistant from the points Q and R
This means that
PQ = PR
These lengths can be calculated using the following distance formula
distance = √[(x₂ - x₁)² + (y₂ - y₁)²]
So, we have
PQ = √[(x - 3)² + (3 - 5)²]
PR = √[(x + 4)² + (3 + 2)²]
Recall that PQ = PR
So, we have
√[(x - 3)² + (3 - 5)²] = √[(x + 4)² + (3 + 2)²]
Square both sides
(x - 3)² + (3 - 5)² = (x + 4)² + (3 + 2)²
So, we have
(x - 3)² + 4 = (x + 4)² + 25
Open the brackets
x² - 6x + 9 + 4 = x² + 8x + 16 + 25
Evaluate the like terms
8x + 6x = 9 + 4 - 16- 25
Evaluate the like terms
14x = -28
This gives
x = -2
Hence, the value of x is -2
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