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
PQ2 = 52 + 42
QR2 = (x-1)2 + 5^2
Looks like x-1 = 4 or x=5
or, x-1 = -4 making x=-3
To find the value of x such that PQ = QR, we need to calculate the lengths of the line segments PQ and QR and equate them.
Let's start by calculating the length of PQ:
PQ = sqrt((x2 - x1)^2 + (y2 - y1)^2)
PQ = sqrt((1 - 6)^2 + (3 - (-1))^2)
PQ = sqrt((-5)^2 + (4)^2)
PQ = sqrt(25 + 16)
PQ = sqrt(41)
Similarly, let's calculate the length of QR:
QR = sqrt((x2 - x1)^2 + (y2 - y1)^2)
QR = sqrt((x - 1)^2 + (8 - 3)^2)
QR = sqrt((x - 1)^2 + (5)^2)
QR = sqrt(x^2 - 2x + 1 + 25)
QR = sqrt(x^2 - 2x + 26)
Now, we set up the equation:
sqrt(41) = sqrt(x^2 - 2x + 26)
To eliminate the square root, we can square both sides of the equation:
41 = x^2 - 2x + 26
Rearranging the equation:
x^2 - 2x - 15 = 0
Now, we can solve this quadratic equation by factoring or applying the quadratic formula. Factoring the equation, we have:
(x - 5)(x + 3) = 0
Setting each factor equal to zero, we get:
x - 5 = 0 --> x = 5
x + 3 = 0 --> x = -3
Therefore, the value of x such that PQ = QR is x = 5 or x = -3.