Question

Find the value of x such that PQ = QR where the coordinates of P, Q and R are (6, -1), (1, 3) and (x, 8) respectively.

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Answer 1

Answer:

x =5 or  x = -3

Step-by-step explanation:

Answer 2

Step-by-step explanation:

PQ = √[(1-6)^2 + (3+1)^2]

= √ 25+ 16

= √ 41

QR = √[( x-1)^2 + (8-3)^2]

= √[ x^2 + 1 - 2x + 25]

= √[x^2 - 2x + 26]

QR = PQ

x^2 - 2x + 26 = 41

x^2 - 2x + 15= 0

x^2 +5x - 3x +15 = 0

x ( x +5) -3( x+5)

(x -3)(x+5)

x-3 = 0

x = 3

x+5 = 0

x= -5