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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.
Solution
The distance d between two points `(x_1,y_1)` and `(x_2, y_2)` is given by the formula
`d= sqrt((x_1- x_2)^2 + (y_1 - y_2)^2)`
The three given points are P(6,−1), Q(1,3) and R(x,8).
Now let us find the distance between 'P’ and ‘Q’.
`PQ = sqrt((6 - 1)^2 + (-1 -3)^2)`
`= sqrt((5)^2 + (-4)^2)`
`= sqrt(25 + 16)`
`PQ = sqrt(41)`
Now, let us find the distance between 'Q' and 'R'.
`QR = sqrt((1- x)^2 + (3 - 8)^2)`
`QR = sqrt((1 - x)^2 = (-5)^2)`
It is given that both these distances are equal. So, let us equate both the above equations,
PQ = QR
`sqrt(41) =sqrt((1- x)^2 + (-5)^2)`
Squaring on both sides of the equation we get,
`41 = (1 - x^2) + (-5)^2`
`41 = 1 + x^2 - 2x + 25`
`15 = x^2 - 2x`
Now we have a quadratic equation. Solving for the roots of the equation we have,
`x^2 - 2x - 15 = 0`
`x^2 - 5x + 3x - 15 = 0`
`x(x - 5) + 3(x - 5) = 0`
(x - 5)(x + 3) = 0
Thus the roots of the above equation are 5 and −3.
Hence the values of 'x' are 5 or -3
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