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.

Answer 1

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