Question

consider reflections of △jkl. what line of reflection maps point k to point k at (-5, 2)? x - axis what line of reflection maps point l to point l at (-2, 3)?

Explanation:

planation:

Step1: Find mid - point of L and L'

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For $L(-3,2)$ and $L'(-2,3)$, the mid - point is $(\frac{-3+( - 2)}{2},\frac{2 + 3}{2})=(-\frac{5}{2},\frac{5}{2})$.

Step2: Find slope of line passing through L and L'

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For $L(-3,2)$ and $L'(-2,3)$, the slope $m=\frac{3 - 2}{-2-( - 3)} = 1$.

Step3: Find slope of the line of reflection

The line of reflection is perpendicular to the line joining the point and its image. If two lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1m_2=-1$. Since the slope of the line joining $L$ and $L'$ is $1$, the slope of the line of reflection is $-1$.

Step4: Find equation of the line of reflection

Using the point - slope form of a line $y - y_0=m(x - x_0)$ with the mid - point $(-\frac{5}{2},\frac{5}{2})$ and slope $-1$, we have $y-\frac{5}{2}=-1(x+\frac{5}{2})$. Simplifying gives $y=-x$.

Answer:

$y = -x$