Answer:
B = (3, -5)
D = (0, -1)
C = (3, 5)
Explanation:
We want to find the coordinates of B, C, and D given that AB = 5 and BC = 10.
Because AB = 5 and BC = 10, we can find the slope of the line using the slope formula:
![\displaystyle \begin{aligned} m & = (\Delta y)/(\Delta x) \\ \\ & = ((10))/((5)) \\ \\ & = 2 \end{aligned}](https://img.qammunity.org/2023/formulas/mathematics/high-school/sycdukqsv0ow1rccmzj874ptodnyibf2b4.png)
Point A(-2,-5) is on the line. Hence, the equation of the line is:
![\displaystyle \begin{aligned} y-y_1 &= m(x-x_1) \\ \\& = y-(-5) = 2(x-(-2)) \\ \\ y + 5 & = 2x+4 \\ \\ y & = 2x-1 \end{aligned}](https://img.qammunity.org/2023/formulas/mathematics/high-school/hjvq4sm4q8vvh8z7ji0bdkwxoiobbn29ba.png)
Because AB is 5, B is simply A shifted rightwards five units. Hence:
![\displaystyle B = (-2+5, -5) = (3, -5)](https://img.qammunity.org/2023/formulas/mathematics/high-school/ms6npkoau7csompyyh052iabd8zhrpbjmn.png)
At Point D, x = 0. Hence:
![\displaystyle y = 2(0) -1 = -1](https://img.qammunity.org/2023/formulas/mathematics/high-school/6g9brqwto7d6es2wxnt6azle6mmg99flhz.png)
Thus, D is at (0, -1).
B and C are collinear. Hence, the x value of C is 3:
![\displaystyle y = 3(2) - 1 = 5](https://img.qammunity.org/2023/formulas/mathematics/high-school/36yplvnv6c75ebu3tz9px41b26ycqaasy3.png)
Therefore, C is at (3, 5).