Question

on the coordinate plane, the segment from d(-6, 3) to e(2, 3) forms one side of △def. the triangle has an area of 28 square units. select all of the points where f could be. (7, 4) (5, -4) (1, 10) (-6, 10)

Explanation:

Step1: Calculate the length of side DE

The distance formula for two - points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For points $D(-6,3)$ and $E(2,3)$, since $y_1 = y_2=3$, the length of $DE=\vert2-(-6)\vert=8$.

Step2: Use the area formula of a triangle

The area formula of a triangle is $A=\frac{1}{2}\times base\times height$. Here, the base is the length of $DE = 8$, and $A = 28$. Substitute into the formula: $28=\frac{1}{2}\times8\times h$. Solving for $h$ gives $h = 7$.

Step3: Check the distance from each point to the line $y = 3$

The line containing segment $DE$ is $y = 3$.

• For point $(7,4)$: The distance from $(7,4)$ to the line $y = 3$ is $\vert4 - 3\vert=1

eq7$.

• For point $(5,-4)$: The distance from $(5,-4)$ to the line $y = 3$ is $\vert-4 - 3\vert=7$.
• For point $(1,10)$: The distance from $(1,10)$ to the line $y = 3$ is $\vert10 - 3\vert=7$.
• For point $(-6,10)$: The distance from $(-6,10)$ to the line $y = 3$ is $\vert10 - 3\vert=7$.

Answer:

$(5,-4),(1,10),(-6,10)$