SOLUTION: What is the equation of the reflection of the line y = 2x + 1 across the line y = 5x - 3? Give your answer in slope-intercept form.

Question 1210146: What is the equation of the reflection of the line y = 2x + 1 across the line y = 5x - 3? Give your answer in slope-intercept form.

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let $L_1$ be the line $y = 2x + 1$, and let $L_2$ be the line $y = 5x - 3$.
We want to find the equation of the reflection of $L_1$ across $L_2$.
Let $P(x_1, y_1)$ be a point on $L_1$, so $y_1 = 2x_1 + 1$.
Let $P'(x_2, y_2)$ be the reflection of $P$ across $L_2$.
The midpoint of $PP'$ lies on $L_2$, and $PP'$ is perpendicular to $L_2$.
The midpoint of $PP'$ is $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
Since $M$ lies on $L_2$, we have
$$\frac{y_1+y_2}{2} = 5\left(\frac{x_1+x_2}{2}\right) - 3$$
$$y_1+y_2 = 5(x_1+x_2) - 6$$
$$y_1+y_2 = 5x_1+5x_2 - 6$$
Substituting $y_1 = 2x_1 + 1$, we get
$$2x_1+1+y_2 = 5x_1+5x_2 - 6$$
$$y_2 = 3x_1+5x_2 - 7 \quad (*)$$
Since $PP'$ is perpendicular to $L_2$, the slope of $PP'$ is the negative reciprocal of the slope of $L_2$.
The slope of $L_2$ is 5, so the slope of $PP'$ is $-\frac{1}{5}$.
Thus,
$$\frac{y_2-y_1}{x_2-x_1} = -\frac{1}{5}$$
$$5(y_2-y_1) = -(x_2-x_1)$$
$$5y_2 - 5y_1 = -x_2 + x_1$$
Substituting $y_1 = 2x_1 + 1$, we get
$$5y_2 - 5(2x_1+1) = -x_2 + x_1$$
$$5y_2 - 10x_1 - 5 = -x_2 + x_1$$
$$5y_2 = 11x_1 - x_2 + 5 \quad (**)$$
From (*), we have $3x_1 = y_2 - 5x_2 + 7$.
Substituting this into (**), we get
$$5y_2 = \frac{11}{3}(y_2 - 5x_2 + 7) - x_2 + 5$$
$$15y_2 = 11y_2 - 55x_2 + 77 - 3x_2 + 15$$
$$4y_2 = -58x_2 + 92$$
$$y_2 = -\frac{58}{4}x_2 + \frac{92}{4}$$
$$y_2 = -\frac{29}{2}x_2 + 23$$
Thus, the equation of the reflected line is $y = -\frac{29}{2}x + 23$.
Final Answer: The final answer is $\boxed{y = -\frac{29}{2} x + 23}$

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