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
as
at y-axis.
Let the point Q be = ![(0,y)](https://tex.z-dn.net/?f=%280%2Cy%29)
We use the distance formula to find length of PQ.
By distance formula:
The distance between two points
and
is given as:
![d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_1-x_2%29%5E2%2B%28y_1-y_2%29%5E2%7D)
Thus, for the point P(8,11) and Q(0,
) the distance PQ can be given as:
![PQ=\sqrt{(8-0)^2+(11-y)^2}](https://tex.z-dn.net/?f=PQ%3D%5Csqrt%7B%288-0%29%5E2%2B%2811-y%29%5E2%7D)
![PQ=\sqrt{(8)^2+(11-y)^2}](https://tex.z-dn.net/?f=PQ%3D%5Csqrt%7B%288%29%5E2%2B%2811-y%29%5E2%7D)
![PQ=\sqrt{64+(11-y)^2}](https://tex.z-dn.net/?f=PQ%3D%5Csqrt%7B64%2B%2811-y%29%5E2%7D)
Substituting PQ=10 units.
![10=\sqrt{64+(11-y)^2}](https://tex.z-dn.net/?f=10%3D%5Csqrt%7B64%2B%2811-y%29%5E2%7D)
Squaring both sides.
![10^2=(\sqrt{64+(11-y)^2})^2](https://tex.z-dn.net/?f=10%5E2%3D%28%5Csqrt%7B64%2B%2811-y%29%5E2%7D%29%5E2)
![100=64+(11-y)^2](https://tex.z-dn.net/?f=100%3D64%2B%2811-y%29%5E2)
Subtracting both sides by 64.
![100-64=64-64+(11-y)^2](https://tex.z-dn.net/?f=100-64%3D64-64%2B%2811-y%29%5E2)
![36=(11-y)^2](https://tex.z-dn.net/?f=36%3D%2811-y%29%5E2)
Taking square root both sides.
![\sqrt{36}=\sqrt{(11-y)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B36%7D%3D%5Csqrt%7B%2811-y%29%5E2%7D)
![\pm6=11-y](https://tex.z-dn.net/?f=%5Cpm6%3D11-y)
So, we have two equations to solve:
and ![-6=11-y](https://tex.z-dn.net/?f=-6%3D11-y)
Adding
both sides.
and ![-6+y=11-y+y](https://tex.z-dn.net/?f=-6%2By%3D11-y%2By)
and ![-6+y=11](https://tex.z-dn.net/?f=-6%2By%3D11)
Subtracting both sides by 6 for one equation and adding 6 both sides for the other equation.
and ![-6+6+y=11+6](https://tex.z-dn.net/?f=-6%2B6%2By%3D11%2B6)
∴
and
Thus, the possible co-ordinates of the point Q are (0,5) and (0,17).