Question

Draw the line of reflection that reflects △ABC onto triangle Δ A'B'C'

Answer 1

Answer:  Line of reflection is y = 2 x + 2

Step-by-step explanation:

By the given diagram,

A≡(2,1) A'≡(-2,3), B≡(5,2) and B'≡(-3,6),

Since, the midpoint of the corresponding vertices of a figure and its reflected figure must lie on the line of reflection.

Here, Δ ABC is reflected onto triangle Δ A'B'C' through this line,

Let, P is the mid point of AA'

⇒ $\text{Coordinates of P}= (\frac{2+(-2)}{2}, \frac{1+3}{2})=(\frac{2-2}{2}, \frac{4}{2})=(0,2)$

Similarly, Q is the mid point of BB'

⇒ $\text{Coordinates of Q} = (\frac{5+(-3)}{2}, \frac{2+6}{2})=(\frac{5-3}{2}, \frac{8}{2})=(1,4)$

Thus, by the above statement,

P and Q are lie on the line of the reflection,

Thus, the equation of the line of reflection,

$y-2 = \frac{4-2}{1-0} (x-0)$

⇒ $y-2 = \frac{2}{1} (x-0)$

⇒ $y= 2(x-0)+2$

⇒ $y= 2x+2$

Which is the required line of reflection.