# Find the Values of Y for Which the Distance Between the Points P (2, —3) and Q (10, Y) is 10 Units - Mathematics

Find the values of y for which the distance between the points P (2, -3) and Q (10, y) is
10 units

#### 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 distance between two points P(2,3) and Q(10,y) is given as 10 units. Substituting these values in the formula for distance between two points we have,

10 = sqrt((2 - 10)^2 + (-3 - y)^2)

Now, squaring the above equation on both sides of the equals sign

100 = (-8)^2 + (-3 - y)^2

100 = 64 + 9 + y^2 + 6y

27 = y^2 + 6y

Thus we arrive at a quadratic equation. Let us solve this now,

y^2 + 6y - 27 = 0

y^2 + 9y - 3y - 27 = 0

y(y + 9) - 3(y + 9) = 0

(y - 3)(y + 9) = 0

The roots of the above quadratic equation are thus 3 and −9.

Thus the value of ‘y’ could either be 3 or -9

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Chapter 6: Co-Ordinate Geometry - Exercise 6.2 [Page 16]

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.2 | Q 35 | Page 16

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