# If the Distance Between the Points (3, 0) and (0, Y) is 5 Units and Y is Positive. Then What is the Value of Y? - Mathematics

Short Note

If the distance between the points (3, 0) and (0, y) is 5 units and y is positive. then what is the value of y?

#### Solution

It is given that distance between P (3, 0) and Q (0 , y)  is 5.

In general, the distance between A(x_1 , y_1 ) "  and B "(x_2 , y_12)  is given by,

AB^2 = (x_2 - x_1) ^2 + ( y_2 - y_1)^2

So,

5^2 = (0 -3)^2 + ( y - 0)^2

On further simplification,

y^2 = 16

 y = +-4

We will neglect the negative value. So,

y = 4

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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Q 24 | Page 62