Question

P is a point (8,11). Q is a point on the y-axis so that PQ=10. Find the coordinates of Q.

Answer 1

The possible co-ordinates of the point Q are (0,5) and (0,17).

Explanation:

Given:

P is a point (8,11)

Q is point on y-axis

PQ = 10 units

To find co-ordinates of point Q.

Solution:

Any point on y-axis is given as
`(0,y)` as
`x=0` at y-axis.

Let the point Q be =
`(0,y)`

We use the distance formula to find length of PQ.

By distance formula:

The distance between two points
`(x_1,y_1)` and
`(x_2,y_2)` is given as:

`d=√((x_1-x_2)^2+(y_1-y_2)^2)`

Thus, for the point P(8,11) and Q(0,
`y`) the distance PQ can be given as:

`PQ=√((8-0)^2+(11-y)^2)`

`PQ=√((8)^2+(11-y)^2)`

`PQ=√(64+(11-y)^2)`

Substituting PQ=10 units.

`10=√(64+(11-y)^2)`

Squaring both sides.

`10^2=(√(64+(11-y)^2))^2`

`100=64+(11-y)^2`

Subtracting both sides by 64.

`100-64=64-64+(11-y)^2`

`36=(11-y)^2`

Taking square root both sides.

`√(36)=√((11-y)^2)`

`\pm6=11-y`

So, we have two equations to solve:

`6=11-y` and
`-6=11-y`

Adding
`y` both sides.

`6+y=11-y+y` and
`-6+y=11-y+y`

`6+y=11` and
`-6+y=11`

Subtracting both sides by 6 for one equation and adding 6 both sides for the other equation.

`6-6+y=11-6` and
`-6+6+y=11+6`

∴
`y=5` and
`y=17`

Thus, the possible co-ordinates of the point Q are (0,5) and (0,17).