Question

which triangle on the coordinate plane is similar to $\triangle a$?

Explanation:

Step1: Identify vertices of △A

Vertices: $(-6,2)$, $(-4,6)$, $(-2,0)$

Step2: Calculate side lengths of △A

Use distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$:

• $d_1=\sqrt{(-4+6)^2+(6-2)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$
• $d_2=\sqrt{(-2+4)^2+(0-6)^2}=\sqrt{4+36}=\sqrt{40}=2\sqrt{10}$
• $d_3=\sqrt{(-2+6)^2+(0-2)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$

Ratio of sides: $2\sqrt{5}:2\sqrt{10}:2\sqrt{5} = 1:\sqrt{2}:1$

Step3: Analyze △B vertices and sides

Vertices: $(2,0)$, $(6,4)$, $(7,2)$

• $d_1=\sqrt{(6-2)^2+(4-0)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$
• $d_2=\sqrt{(7-6)^2+(2-4)^2}=\sqrt{1+4}=\sqrt{5}$
• $d_3=\sqrt{(7-2)^2+(2-0)^2}=\sqrt{25+4}=\sqrt{29}$

Ratio does not match $1:\sqrt{2}:1$

Step4: Analyze △C vertices and sides

Vertices: $(-3,-2)$, $(-1,-1)$, $(-2,-3)$

• $d_1=\sqrt{(-1+3)^2+(-1+2)^2}=\sqrt{4+1}=\sqrt{5}$
• $d_2=\sqrt{(-2+1)^2+(-3+1)^2}=\sqrt{1+4}=\sqrt{5}$
• $d_3=\sqrt{(-2+3)^2+(-3+2)^2}=\sqrt{1+1}=\sqrt{2}$

Ratio of sides: $\sqrt{5}:\sqrt{5}:\sqrt{2}=1:1:\frac{\sqrt{2}}{\sqrt{5}}$ (simplify to $1:1:\frac{\sqrt{10}}{5}$)
Rewrite ratio: $\sqrt{5}:\sqrt{2}:\sqrt{5}=1:\frac{\sqrt{2}}{\sqrt{5}}:1 = 1:\frac{\sqrt{10}}{5}:1$, scale by $\sqrt{5}$: $\sqrt{5}:\sqrt{2}:\sqrt{5}$, matches scaled $1:\sqrt{2}:1$ (scale factor $\frac{1}{\sqrt{5}}$)

Step5: Analyze △D vertices and sides

Vertices: $(1,-1)$, $(2,0)$, $(3,-4)$

• $d_1=\sqrt{(2-1)^2+(0+1)^2}=\sqrt{1+1}=\sqrt{2}$
• $d_2=\sqrt{(3-2)^2+(-4-0)^2}=\sqrt{1+16}=\sqrt{17}$
• $d_3=\sqrt{(3-1)^2+(-4+1)^2}=\sqrt{4+9}=\sqrt{13}$

Ratio does not match $1:\sqrt{2}:1$

Answer:

$\triangle C$