Question

honors algebra 2 mr. koomson homework for 9 - 04 - 25 question#1: find the area of the triangle with the coordinates: a(10, 10), b(2, -2), and c(-5, 5). step#1: choose a side to be your base and find the length of that side. we chose side ab as our base in class. step#2. find the equation of the line ab step#3: find the equation of the line that passes through point c and is perpendicular to line ab. call the point of intersection q. step#4: solve the system of equations (cq vs ab) or use desmos to find the coordinates of q. step#5: use the distance formula to find the length of cq. that is the height of the triangle. you can now find the area of triangle abc using a = ½baseheight question#2: solve the same problem again using a different side of the triangle as your base.

Explanation:

Step1: Calculate length of base AB

The distance formula between 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 $A(10,10)$ and $B(2,-2)$, we have $x_1 = 10,y_1 = 10,x_2 = 2,y_2=-2$. Then $AB=\sqrt{(2 - 10)^2+(-2 - 10)^2}=\sqrt{(-8)^2+(-12)^2}=\sqrt{64 + 144}=\sqrt{208}=4\sqrt{13}$.

Step2: Find equation of line AB

The slope $m$ of line passing through $A(10,10)$ and $B(2,-2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-2 - 10}{2 - 10}=\frac{-12}{-8}=\frac{3}{2}$. Using the point - slope form $y - y_1=m(x - x_1)$ with point $A(10,10)$, we get $y - 10=\frac{3}{2}(x - 10)$, which simplifies to $y=\frac{3}{2}x - 5$.

Step3: Find equation of line perpendicular to AB through C

The slope of a line perpendicular to a line with slope $m$ is $-\frac{2}{3}$ (since $m_1\times m_2=-1$ for perpendicular lines). Using the point - slope form with $C(-5,5)$, we have $y - 5=-\frac{2}{3}(x + 5)$, which simplifies to $y=-\frac{2}{3}x+\frac{5}{3}$.

Step4: Find intersection point Q

We solve the system of equations

$$\begin{cases}y=\frac{3}{2}x - 5\\y=-\frac{2}{3}x+\frac{5}{3}\end{cases}$$

. Set $\frac{3}{2}x - 5=-\frac{2}{3}x+\frac{5}{3}$. Multiply through by 6 to clear the fractions: $9x-30=-4x + 10$. Combine like terms: $9x + 4x=10 + 30$, so $13x=40$, and $x=\frac{40}{13}$. Substitute $x=\frac{40}{13}$ into $y=\frac{3}{2}x - 5$ gives $y=\frac{3}{2}\times\frac{40}{13}-5=\frac{60}{13}-\frac{65}{13}=-\frac{5}{13}$. So $Q(\frac{40}{13},-\frac{5}{13})$.

Step5: Calculate height CQ

Using the distance formula between $C(-5,5)$ and $Q(\frac{40}{13},-\frac{5}{13})$, $CQ=\sqrt{(-5-\frac{40}{13})^2+(5+\frac{5}{13})^2}=\sqrt{(-\frac{65 + 40}{13})^2+(\frac{65+5}{13})^2}=\sqrt{(-\frac{105}{13})^2+(\frac{70}{13})^2}=\sqrt{\frac{11025+4900}{13^2}}=\sqrt{\frac{15925}{13^2}}=\frac{5\sqrt{637}}{13}$.

Step6: Calculate area

$A=\frac{1}{2}\times AB\times CQ=\frac{1}{2}\times4\sqrt{13}\times\frac{5\sqrt{637}}{13}=\frac{1}{2}\times4\sqrt{13}\times\frac{5\times7\sqrt{13}}{13}=\frac{1}{2}\times4\times5\times7 = 70$.

Answer:

The area of triangle $ABC$ is 70.

For Question#2, we can choose side $BC$ or $AC$ as the base and follow the same steps as above to re - calculate the area. The result should be the same 70.