Question

ONM has vertices O (-4, 2), N (3, 6) and M(0,3). What are the coordinates of the

vertices of O' N' M' if ONM is first translated (x, y)+(x + 3, y) followed by a

reflection across the line y = 1?

Answer 1

Answer:

The new vertices are $O''(x,y) = (-1, 0)$, $N''(x,y) = (6,-4)$ and $M''(x,y) = (3,-1)$.

Step-by-step explanation:

Previously, we have to define a vectorial equation for a reflection across the line $y = 1$:

Reflection across the line y = 1:

$P'(x,y) = P(x,y) -2\cdot [P(x,y) -(x_{P},1)]$ (1)

Where:

$P(x,y)$ - Original point.

$x_{P}$ - x-Coordinate of the original point.

$P'(x,y)$ - Resulting point.

Translation:

$P'(x,y) = P(x,y) + (3,0)$ (2)

If we know that $O(x,y) = (-4,2)$, $N(x,y) = (3, 6)$ and $M(x,y) = (0,3)$, then the resulting points are, respectively:

Point O'

Translation

$O'(x,y) = O(x,y) + (3,0)$

$O'(x,y) = (-1,2)$

Reflection

$O''(x,y) = O'(x,y) -2\cdot [O'(x,y) -(x_{O'},1)]$

$O''(x,y) = (-1,2) -2\cdot [(-1, 2)-(-1,1)]$

$O''(x,y) = (-1, 0)$

Point N'

Translation

$N'(x,y) = N(x,y) + (3,0)$

$N'(x,y) = (6,6)$

Reflection

$N''(x,y) = N'(x,y) -2\cdot [N'(x,y) -(x_{N'},1)]$

$N''(x,y) = (6,6) - 2\cdot [(6,6)-(6,1)]$

$N''(x,y) = (6,-4)$

Point M'

Translation

$M'(x,y) = M(x,y) + (3,0)$

$M(x,y) = (3,3)$

Reflection

$M''(x,y) = M'(x,y) -2\cdot [M'(x,y) -(x_{M'},1)]$

$M''(x,y) = (3,3) - 2\cdot [(3,3)-(3,1)]$

$M''(x,y) = (3,-1)$

The new vertices are $O''(x,y) = (-1, 0)$, $N''(x,y) = (6,-4)$ and $M''(x,y) = (3,-1)$.