# 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. - Mathematics

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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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.2 | Q 31 | Page 16