Question

find the area, in square units, of △abc plotted below.
a(0,7)
b(7, -2)
c(-3, -4)
d(2, -3)

Explanation:

Step1: Find length of base BC

Use distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. For $B(7, -2)$ and $C(-3, -4)$:

$$\begin{align*} BC&=\sqrt{(7 - (-3))^2 + (-2 - (-4))^2}\\ &=\sqrt{(10)^2 + (2)^2}\\ &=\sqrt{100 + 4}\\ &=\sqrt{104}\\ &=2\sqrt{26} \end{align*}$$

Wait, alternatively, since D is on BC, maybe use horizontal/vertical? Wait, no, better to find the length of BC by x - difference: from x=-3 to x=7, that's 10 units? Wait no, y - coordinates: C(-3,-4), B(7,-2). Wait, maybe the base is BC, and the height is the vertical distance from A to the line BC? Wait, no, the dashed line is from A to D(2,-3), which is perpendicular? Wait, maybe first find the equation of line BC, then find the distance from A to BC (height), then use area = ½baseheight.

First, find slope of BC: $m_{BC}=\frac{-2 - (-4)}{7 - (-3)}=\frac{2}{10}=\frac{1}{5}$. Equation of BC: using point C(-3,-4): $y - (-4)=\frac{1}{5}(x - (-3))$ → $y + 4=\frac{1}{5}(x + 3)$ → $5y + 20 = x + 3$ → $x - 5y - 17 = 0$.

Distance from A(0,7) to line BC (height h): formula $d=\frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}$. Line BC: $x - 5y - 17 = 0$, so A=1, B=-5, C=-17.
$$h=\frac{|1*0 + (-5)*7 - 17|}{\sqrt{1^2 + (-5)^2}}=\frac{|-35 - 17|}{\sqrt{26}}=\frac{52}{\sqrt{26}}=2\sqrt{26}$$

Length of BC: distance between B(7,-2) and C(-3,-4):
$$BC=\sqrt{(7 - (-3))^2 + (-2 - (-4))^2}=\sqrt{10^2 + 2^2}=\sqrt{104}=2\sqrt{26}$$

Then area = ½baseheight = ½2√262√26 = ½426 = 52. Wait, that's complicated. Alternatively, use the shoelace formula.

Shoelace formula for triangle with vertices (x1,y1), (x2,y2), (x3,y3):
Area = ½|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

For A(0,7), B(7,-2), C(-3,-4):

$$\begin{align*} \text{Area}&=\frac{1}{2}|0*(-2 - (-4)) + 7*(-4 - 7) + (-3)*(7 - (-2))|\\ &=\frac{1}{2}|0 + 7*(-11) + (-3)*(9)|\\ &=\frac{1}{2}| -77 - 27 |\\ &=\frac{1}{2}| -104 |\\ &=\frac{1}{2}*104\\ &=52 \end{align*}$$

Step2: Verify with base and height (simpler way)

Wait, looking at the graph, D is (2,-3). Let's check the length of BC: from x=-3 to x=7, that's 10 units? Wait no, x-coordinates: C(-3,-4), B(7,-2). Wait, maybe the base is along the x - direction? Wait, no, the y - coordinates of C and B are -4 and -2, close. Wait, alternatively, the base is BC, and the height is the vertical distance from A to the line BC? No, better shoelace is easier.

Answer:

52