A triangle has vertices $$ A(2,-2), B(6,4) $$ and $$ C(10,1) $$. Determine the equation of the sides of the triangle. A parallelogram has vertices $$ A(2,1), B(5,4), C(9,4) $$ and $$ D(6,1) $$. Determine the equation of line: $$ \begin{array}{llll} ext { a) } A B & ext { b) } B C & ext { c) } C D & ext { d) } A D\end{array} $$

Question

A triangle has vertices $$ A(2,-2), B(6,4) $$ and $$ C(10,1) $$. Determine the equation of the sides of the triangle. A parallelogram has vertices $$ A(2,1), B(5,4), C(9,4) $$ and $$ D(6,1) $$. Determine the equation of line: $$ \begin{array}{llll}	ext { a) } A B & 	ext { b) } B C & 	ext { c) } C D & 	ext { d) } A D\end{array} $$

UpStudy AI · Accepted Answer

**Triangle**

Let the vertices be  
$$ A(2,-2) $$, $$ B(6,4) $$, and $$ C(10,1) $$.

1. **Side $$ AB $$** (through $$ A $$ and $$ B $$)  
   - Calculate the slope:  
     $$
     m_{AB}=\frac{4-(-2)}{6-2}=\frac{6}{4}=\frac{3}{2}
     $$
   - Write the point-slope form using point $$ A(2,-2) $$:  
     $$
     y-(-2)=\frac{3}{2}(x-2)
     $$
     This simplifies to:  
     $$
     y+2=\frac{3}{2}(x-2)
     $$
   - Solve for $$ y $$:  
     $$
     y=\frac{3}{2}x-3-2=\frac{3}{2}x-5
     $$
     
2. **Side $$ BC $$** (through $$ B $$ and $$ C $$)  
   - Calculate the slope:  
     $$
     m_{BC}=\frac{1-4}{10-6}=\frac{-3}{4}=-\frac{3}{4}
     $$
   - Use point $$ B(6,4) $$:  
     $$
     y-4=-\frac{3}{4}(x-6)
     $$
   - Solve for $$ y $$:  
     $$
     y=-\frac{3}{4}x+\frac{3}{4}\cdot6+4=-\frac{3}{4}x+ \frac{18}{4}+4=-\frac{3}{4}x+\frac{9}{2}+4
     $$
     Write $$4$$ as $$\frac{8}{2}$$:  
     $$
     y=-\frac{3}{4}x+\frac{9}{2}+\frac{8}{2}=-\frac{3}{4}x+\frac{17}{2}
     $$
     
3. **Side $$ CA $$** (through $$ C $$ and $$ A $$)  
   - Calculate the slope:  
     $$
     m_{CA}=\frac{-2-1}{2-10}=\frac{-3}{-8}=\frac{3}{8}
     $$
   - Use point $$ C(10,1) $$:  
     $$
     y-1=\frac{3}{8}(x-10)
     $$
   - Solve for $$ y $$:  
     $$
     y=\frac{3}{8}x-\frac{30}{8}+1=\frac{3}{8}x-\frac{30}{8}+\frac{8}{8}=\frac{3}{8}x-\frac{22}{8}=\frac{3}{8}x-\frac{11}{4}
     $$

**Parallelogram**

Let the vertices be  
$$ A(2,1) $$, $$ B(5,4) $$, $$ C(9,4) $$, and $$ D(6,1) $$.

1. **Line $$ AB $$** (through $$ A $$ and $$ B $$)  
   - Slope:  
     $$
     m_{AB}=\frac{4-1}{5-2}=\frac{3}{3}=1
     $$
   - Equation using point $$ A(2,1) $$:  
     $$
     y-1=1\cdot (x-2)
     $$
     So,  
     $$
     y=x-1
     $$

2. **Line $$ BC $$** (through $$ B $$ and $$ C $$)  
   - Slope:  
     $$
     m_{BC}=\frac{4-4}{9-5}=\frac{0}{4}=0
     $$
   - The line is horizontal, so:  
     $$
     y=4
     $$

3. **Line $$ CD $$** (through $$ C $$ and $$ D $$)  
   - Slope:  
     $$
     m_{CD}=\frac{1-4}{6-9}=\frac{-3}{-3}=1
     $$
   - Using point $$ C(9,4) $$:  
     $$
     y-4=1\cdot (x-9)
     $$
     Thus,  
     $$
     y=x-5
     $$

4. **Line $$ AD $$** (through $$ A $$ and $$ D $$)  
   - Slope:  
     $$
     m_{AD}=\frac{1-1}{6-2}=\frac{0}{4}=0
     $$
   - The line is horizontal:  
     $$
     y=1
     $$
**Triangle:** - **Side $$ AB $$:** $$ y = \frac{3}{2}x - 5 $$ - **Side $$ BC $$:** $$ y = -\frac{3}{4}x + \frac{17}{2} $$ - **Side $$ CA $$:** $$ y = \frac{3}{8}x - \frac{11}{4} $$ **Parallelogram:** - **Line $$ AB $$:** $$ y = x - 1 $$ - **Line $$ BC $$:** $$ y = 4 $$ - **Line $$ CD $$:** $$ y = x - 5 $$ - **Line $$ AD $$:** $$ y = 1 $$

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