Question

plot and connect the points listed below. when you are done, find the area of the triangle.
a(-3, 4), b(2, 2), c(-3, 0)
click on the graph to plot a point. click a point to delete it.
when you are done, press \graph figure\ to see the result.
graph with x-axis from -6 to 6 and y-axis from -5 to 6, origin at (0,0)

Explanation:

Step1: Identify the base and height

Points A(-3, 4) and C(-3, 0) have the same x - coordinate, so the length of AC (the base) is the difference in y - coordinates.
Length of \( AC=\vert4 - 0\vert=4\).
The horizontal distance from point B(2, 2) to the line \( x=-3\) (where points A and C lie) is the absolute difference between the x - coordinates of B and the line \( x = - 3\). So the height \( h=\vert2-(-3)\vert=\vert2 + 3\vert = 5\)? Wait, no. Wait, for a triangle with a vertical side (AC is vertical), the base is AC and the height is the horizontal distance from B to the line AC. Wait, actually, the formula for the area of a triangle is \( A=\frac{1}{2}\times base\times height\). Since AC is vertical (x=-3 for both A and C), the length of AC is \( y_A - y_C=4 - 0 = 4\) (since \( y_A>y_C\)). The horizontal distance from point B(2,2) to the line \( x=-3\) is \( 2-(-3)=5\)? Wait, no, the base is AC, and the height is the horizontal distance from B to the line AC. Wait, actually, when we have a vertical line segment AC, the base is AC, and the height is the horizontal distance between the line AC (x = - 3) and the x - coordinate of B (x = 2). So the horizontal distance is \( 2-(-3)=5\)? Wait, no, that's not right. Wait, let's re - examine the coordinates.

Points A(-3,4), C(-3,0): the distance between A and C is \( \sqrt{(-3 + 3)^2+(4 - 0)^2}=\sqrt{0 + 16}=4\). So the length of AC is 4. Now, the line AC is a vertical line (x=-3). The point B is at (2,2). The height of the triangle with respect to base AC is the horizontal distance from B to the line AC. The horizontal distance between x = 2 and x=-3 is \( \vert2-(-3)\vert=\vert2 + 3\vert=5\)? Wait, no, that can't be. Wait, no, actually, when the base is vertical, the height is the horizontal distance, but let's use the formula for the area of a triangle given coordinates. The formula for the area of a triangle with vertices \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\) is also given by the shoelace formula: \( A=\frac{1}{2}\vert x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)\vert\).

Substituting \( x_1=-3,y_1 = 4\); \( x_2=2,y_2 = 2\); \( x_3=-3,y_3 = 0\) into the shoelace formula:

\( A=\frac{1}{2}\vert-3(2 - 0)+2(0 - 4)+(-3)(4 - 2)\vert\)

\(=\frac{1}{2}\vert-3\times2+2\times(-4)+(-3)\times2\vert\)

\(=\frac{1}{2}\vert-6-8 - 6\vert=\frac{1}{2}\vert-20\vert = 10\). Wait, that's different from my previous wrong calculation. Let's go back.

Wait, the base is AC, length 4. The height is the horizontal distance from B to the line AC. Wait, no, the line AC is x=-3. The x - coordinate of B is 2. The horizontal distance is \( 2-(-3)=5\)? But then \( \frac{1}{2}\times4\times5 = 10\), which matches the shoelace formula.

Wait, another way: plot the points. A(-3,4), C(-3,0) are on x=-3, 4 units apart vertically. B(2,2) is to the right of x=-3. The triangle is formed by connecting these three points. The base is AC = 4, and the height is the horizontal distance from B to the line AC, which is \( 2-(-3)=5\). Then area \( A=\frac{1}{2}\times base\times height=\frac{1}{2}\times4\times5 = 10\).

Step2: Calculate the area

Using the formula for the area of a triangle \( A=\frac{1}{2}\times base\times height\). We found that the base (length of AC) is 4 and the height (horizontal distance from B to AC) is 5.

So \( A=\frac{1}{2}\times4\times5\)

\(=\frac{1}{2}\times20 = 10\)

Answer:

10