Given:
Point B has coordinates (4,1).
The x-coordinate of point A is -4.
The distance between point A and point B is 10 units.
To find:
The possible coordinates of point A.
Solution:
Let the y-coordinate of point A be y. Then the two points are A(-4,y) and B(4,1).
Distance formula:
![D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=D%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
The distance between point A and point B is 10 units.
![\sqrt{(4-(-4))^2+(1-y)^2}=10](https://tex.z-dn.net/?f=%5Csqrt%7B%284-%28-4%29%29%5E2%2B%281-y%29%5E2%7D%3D10)
Taking square on both sides, we get
![(8)^2+(1-y)^2=100](https://tex.z-dn.net/?f=%288%29%5E2%2B%281-y%29%5E2%3D100)
![(1-y)^2=100-64](https://tex.z-dn.net/?f=%281-y%29%5E2%3D100-64)
![(1-y)^2=36](https://tex.z-dn.net/?f=%281-y%29%5E2%3D36)
Taking square root on both sides, we get
![(1-y)=\pm \sqrt{36}](https://tex.z-dn.net/?f=%281-y%29%3D%5Cpm%20%5Csqrt%7B36%7D)
![-y=\pm 6-1](https://tex.z-dn.net/?f=-y%3D%5Cpm%206-1)
![y=1\mp 6](https://tex.z-dn.net/?f=y%3D1%5Cmp%206)
and ![y=1+6](https://tex.z-dn.net/?f=y%3D1%2B6)
and ![y=7](https://tex.z-dn.net/?f=y%3D7)
Therefore, the possible coordinates of point A are either (-4,-5) or (-4,7).