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Question
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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