Question

on the coordinate plane, the segment from a(-4, -3) to b(6, -3) forms one side of △abc. the triangle has an area of 25 square units. select all of the points where c could be.
(9, 1)
(-1, -8)
(-1, 2)
(-1, 1)

Explanation:

Step1: Calculate the length of side AB

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 $A(-4,-3)$ and $B(6,-3)$, since $y_1 = y_2=-3$, the length of $AB$ is $|6-(-4)|=10$.

Step2: Use the triangle - area formula

The area formula of a triangle is $A=\frac{1}{2}\times base\times height$. Here, the base is the length of $AB = 10$, and $A = 25$. Substitute into the formula: $25=\frac{1}{2}\times10\times h$. Solving for $h$ gives $h = 5$.

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

The line containing side $AB$ is $y = - 3$.

• For point $(9,1)$: The distance from the point $(9,1)$ to the line $y=-3$ is $|1-(-3)|=4

eq5$.

• For point $(-1,-8)$: The distance from the point $(-1,-8)$ to the line $y=-3$ is $|-3-(-8)| = 5$.
• For point $(-1,2)$: The distance from the point $(-1,2)$ to the line $y=-3$ is $|2-(-3)|=5$.
• For point $(-1,1)$: The distance from the point $(-1,1)$ to the line $y=-3$ is $|1-(-3)|=4

eq5$.

Answer:

$(-1,-8),(-1,2)$