Let y represent the y-coordinate of point A.
We have been given that point B has coordinates (5,1) The x-coordinate of point A is 0. So coordinates of point A would be (0,y)
The distance between point A and Point B is 13 units.
We will use distance formula to solve our given problem.
![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)
Let point A
and point A
.
Upon substituting coordinates of both points in distance formula, we will get:
![13=\sqrt{(0-5)^2+(y-1)^2}](https://tex.z-dn.net/?f=13%3D%5Csqrt%7B%280-5%29%5E2%2B%28y-1%29%5E2%7D)
![13=\sqrt{25+y^2-2y+1}](https://tex.z-dn.net/?f=13%3D%5Csqrt%7B25%2By%5E2-2y%2B1%7D)
![13=\sqrt{y^2-2y+26}](https://tex.z-dn.net/?f=13%3D%5Csqrt%7By%5E2-2y%2B26%7D)
Let us square both sides as:
![13^2=(\sqrt{y^2-2y+26})^2](https://tex.z-dn.net/?f=13%5E2%3D%28%5Csqrt%7By%5E2-2y%2B26%7D%29%5E2)
![169=y^2-2y+26](https://tex.z-dn.net/?f=169%3Dy%5E2-2y%2B26)
![169-169=y^2-2y+26-169](https://tex.z-dn.net/?f=169-169%3Dy%5E2-2y%2B26-169)
![0=y^2-2y-143](https://tex.z-dn.net/?f=0%3Dy%5E2-2y-143)
![y^2-2y-143=0](https://tex.z-dn.net/?f=y%5E2-2y-143%3D0)
Upon splitting the middle term, we will get:
![y^2-13y+11y-143=0](https://tex.z-dn.net/?f=y%5E2-13y%2B11y-143%3D0)
![y(y-13)+11(y-13)=0](https://tex.z-dn.net/?f=y%28y-13%29%2B11%28y-13%29%3D0)
![(y-13)(y+11)=0](https://tex.z-dn.net/?f=%28y-13%29%28y%2B11%29%3D0)
Now we will use zero product property.
![(y-13)=0, (y+11)=0](https://tex.z-dn.net/?f=%28y-13%29%3D0%2C%20%28y%2B11%29%3D0)
![y=13, y=-11](https://tex.z-dn.net/?f=y%3D13%2C%20y%3D-11)
Therefore, the possible coordinates of point A would be
and
.
Is it 7.1 or something I don’t know for sure
Answer:
5:3
Step-by-step explanation:
10 and 6 can both be simplified to 5 and 3. they wanted strawberries first so you put the 5 first then the 3
Answer:
22
Step-by-step explanation: