Question

find the area, in square units, of $\triangle abc$ plotted below.
$a(-6, 4)$, $d(2, 6)$, $b(6, 7)$, $c(4, -2)$ (with the triangle and grid plotted as shown in the image)

Explanation:

Step1: Find length of AB

Use distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. For A(-6,4) and B(6,7):
$AB = \sqrt{(6 - (-6))^2 + (7 - 4)^2} = \sqrt{12^2 + 3^2} = \sqrt{144 + 9} = \sqrt{153} = 3\sqrt{17}$? Wait, no, wait the diagram has D(2,6) with a right angle. Maybe better to find length of AB using horizontal/vertical? Wait, A(-6,4), B(6,7). Wait, no, the height is from C to AB, with D on AB. Let's find length of AB first. Wait, coordinates: A(-6,4), B(6,7). The vector AB is (12, 3). The length of AB: $\sqrt{(6 - (-6))^2 + (7 - 4)^2} = \sqrt{144 + 9} = \sqrt{153} = 3\sqrt{17}$? No, wait, maybe the base is AB, and height is the distance from C to line AB. Alternatively, use the formula for area of triangle with coordinates: shoelace formula.

Shoelace formula: For points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$, area is $\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.

Plug in A(-6,4), B(6,7), C(4,-2):

Area = $\frac{1}{2}| -6(7 - (-2)) + 6(-2 - 4) + 4(4 - 7) |$
= $\frac{1}{2}| -6(9) + 6(-6) + 4(-3) |$
= $\frac{1}{2}| -54 - 36 - 12 |$
= $\frac{1}{2}| -102 |$
= $\frac{1}{2} \times 102 = 51$.

Wait, let's check with base and height. Let's find length of AB: distance between A(-6,4) and B(6,7): $\sqrt{(12)^2 + (3)^2} = \sqrt{144 + 9} = \sqrt{153} = 3\sqrt{17}$. Now find the distance from C(4,-2) to line AB.

Equation of line AB: slope $m = \frac{7 - 4}{6 - (-6)} = \frac{3}{12} = \frac{1}{4}$. Equation: $y - 4 = \frac{1}{4}(x + 6)$ → $y = \frac{1}{4}x + \frac{6}{4} + 4$ → $y = \frac{1}{4}x + \frac{3}{2} + 4$ → $y = \frac{1}{4}x + \frac{11}{2}$.

Distance from (4,-2) to this line: $| \frac{1}{4}(4) - 1(-2) + \frac{11}{2} | / \sqrt{(\frac{1}{4})^2 + (-1)^2}$? Wait, no, standard distance formula: $|Ax + By + C| / \sqrt{A^2 + B^2}$ for line $Ax + By + C = 0$.

Rewrite line AB: $x - 4y + 22 = 0$? Wait, $y = \frac{1}{4}x + \frac{11}{2}$ → $x - 4y + 22 = 0$? Let's check: $x - 4y = -22$ → $4y = x + 22$ → $y = \frac{1}{4}x + \frac{22}{4} = \frac{1}{4}x + \frac{11}{2}$. Correct.

Distance from (4,-2) to $x - 4y + 22 = 0$: $|4 - 4(-2) + 22| / \sqrt{1 + 16} = |4 + 8 + 22| / \sqrt{17} = |34| / \sqrt{17} = 34 / \sqrt{17} = 2\sqrt{17}$.

Now area = $\frac{1}{2} \times base \times height = \frac{1}{2} \times 3\sqrt{17} \times 2\sqrt{17} = \frac{1}{2} \times 6 \times 17 = 51$. Yep, matches shoelace. So the area is 51.

Step1 (Alternative): Shoelace Formula

List coordinates: A(-6,4), B(6,7), C(4,-2), back to A(-6,4).

Compute sum of $x_i y_{i+1}$: (-6)(7) + 6(-2) + 4(4) = -42 -12 +16 = -38.

Compute sum of $y_i x_{i+1}$: 4(6) + 7(4) + (-2)(-6) = 24 +28 +12 = 64.

Subtract: -38 - 64 = -102. Absolute value and half: $\frac{1}{2}|-102| = 51$.

Answer:

51