With reference to the given figure, what has remained the same and what has changed with the reflection? Class 9

With reference to the given figure, what has remained the same and what has changed with the reflection? Class 9

Question 1.

With reference to the given figure, what has remained the same and what has changed with the reflection? Class 9

Answer:

AD = [latex]\sqrt{3^2+4^2}[/latex] = 5 units

DM = [latex]\sqrt{2^2+5^2}[/latex] = [latex]\sqrt{29}[/latex] units

MA = [latex]\sqrt{6^2+2^2}[/latex] = [latex]\sqrt{40}[/latex] units

C' D' = x-coordinate of C' - x-coordinate of D' = -3 - (-7) = 4

A'C' = y-coordinate of A' - y-coordinate of C' = 4 - 1 = 3

A'D' = [latex]\sqrt{3^2+4^2}[/latex] =5 units

D'M' = [latex]\sqrt{(-2)^2+5^2}[/latex] = [latex]\sqrt{29}[/latex] units

M'A' = [latex]\sqrt{(-6)^2+2^2}[/latex] = [latex]\sqrt{40}[/latex] units

∴ The lengths of the triangle has remained the same and the coordinates of the vertices of the triangle have changed after reflection.