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

Answer:

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

Step-by-step 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:

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

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

$PQ=\sqrt{(8-0)^2+(11-y)^2}$

$PQ=\sqrt{(8)^2+(11-y)^2}$

$PQ=\sqrt{64+(11-y)^2}$

Substituting PQ=10 units.

$10=\sqrt{64+(11-y)^2}$

Squaring both sides.

$10^2=(\sqrt{64+(11-y)^2})^2$

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

Subtracting both sides by 64.

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

$36=(11-y)^2$

Taking square root both sides.

$\sqrt{36}=\sqrt{(11-y)^2}$

$\pm6=11-y$

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).