QUESTION IMAGE
Question
reflect the figure over the line y = 1/3x + 1. plot all of the points of the reflected figure. you may click a plotted point to delete it.
Step1: Recall reflection formula
The formula to reflect a point $(x_0,y_0)$ over the line $y = mx + c$ is given by:
Let $m=\frac{1}{3}$ and $c = 1$. First, find the perpendicular - distance from the point to the line. The line perpendicular to $y=mx + c$ passing through $(x_0,y_0)$ has a slope $m'=-\frac{1}{m}=-3$. The equation of the perpendicular line passing through $(x_0,y_0)$ is $y - y_0=-3(x - x_0)$ or $y=-3x+(3x_0 + y_0)$.
Step2: Find intersection point
Solve the system of equations
Set $\frac{1}{3}x + 1=-3x+(3x_0 + y_0)$
$\frac{1}{3}x+3x=3x_0 + y_0 - 1$
$\frac{1 + 9}{3}x=3x_0 + y_0 - 1$
$x=\frac{3(3x_0 + y_0 - 1)}{10}$
Substitute $x$ into $y=\frac{1}{3}x + 1$:
$y=\frac{1}{3}\times\frac{3(3x_0 + y_0 - 1)}{10}+1=\frac{3x_0 + y_0 - 1}{10}+1=\frac{3x_0 + y_0 - 1 + 10}{10}=\frac{3x_0 + y_0+9}{10}$
Let the intersection point be $(x_i,y_i)$.
Step3: Use mid - point formula
If $(x_i,y_i)$ is the mid - point between $(x_0,y_0)$ and its reflection $(x_1,y_1)$, then $x_i=\frac{x_0 + x_1}{2}$ and $y_i=\frac{y_0 + y_1}{2}$.
$x_1 = 2x_i-x_0$ and $y_1 = 2y_i-y_0$
Substitute $x_i=\frac{3(3x_0 + y_0 - 1)}{10}$ and $y_i=\frac{3x_0 + y_0+9}{10}$:
$x_1=2\times\frac{3(3x_0 + y_0 - 1)}{10}-x_0=\frac{18x_0 + 6y_0-6 - 10x_0}{10}=\frac{8x_0 + 6y_0-6}{10}=\frac{4x_0 + 3y_0 - 3}{5}$
$y_1=2\times\frac{3x_0 + y_0+9}{10}-y_0=\frac{6x_0 + 2y_0+18 - 10y_0}{10}=\frac{6x_0 - 8y_0+18}{10}=\frac{3x_0 - 4y_0+9}{5}$
For each vertex of the given triangle, use the above formulas to find its reflection and then plot the new points to get the reflected figure.
Since no specific coordinates of the vertices of the triangle are given in text form, we can't give numerical answers. But the general method to find the reflected points of a figure over the line $y=\frac{1}{3}x + 1$ is as above.
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Step1: Recall reflection formula
The formula to reflect a point $(x_0,y_0)$ over the line $y = mx + c$ is given by:
Let $m=\frac{1}{3}$ and $c = 1$. First, find the perpendicular - distance from the point to the line. The line perpendicular to $y=mx + c$ passing through $(x_0,y_0)$ has a slope $m'=-\frac{1}{m}=-3$. The equation of the perpendicular line passing through $(x_0,y_0)$ is $y - y_0=-3(x - x_0)$ or $y=-3x+(3x_0 + y_0)$.
Step2: Find intersection point
Solve the system of equations
Set $\frac{1}{3}x + 1=-3x+(3x_0 + y_0)$
$\frac{1}{3}x+3x=3x_0 + y_0 - 1$
$\frac{1 + 9}{3}x=3x_0 + y_0 - 1$
$x=\frac{3(3x_0 + y_0 - 1)}{10}$
Substitute $x$ into $y=\frac{1}{3}x + 1$:
$y=\frac{1}{3}\times\frac{3(3x_0 + y_0 - 1)}{10}+1=\frac{3x_0 + y_0 - 1}{10}+1=\frac{3x_0 + y_0 - 1 + 10}{10}=\frac{3x_0 + y_0+9}{10}$
Let the intersection point be $(x_i,y_i)$.
Step3: Use mid - point formula
If $(x_i,y_i)$ is the mid - point between $(x_0,y_0)$ and its reflection $(x_1,y_1)$, then $x_i=\frac{x_0 + x_1}{2}$ and $y_i=\frac{y_0 + y_1}{2}$.
$x_1 = 2x_i-x_0$ and $y_1 = 2y_i-y_0$
Substitute $x_i=\frac{3(3x_0 + y_0 - 1)}{10}$ and $y_i=\frac{3x_0 + y_0+9}{10}$:
$x_1=2\times\frac{3(3x_0 + y_0 - 1)}{10}-x_0=\frac{18x_0 + 6y_0-6 - 10x_0}{10}=\frac{8x_0 + 6y_0-6}{10}=\frac{4x_0 + 3y_0 - 3}{5}$
$y_1=2\times\frac{3x_0 + y_0+9}{10}-y_0=\frac{6x_0 + 2y_0+18 - 10y_0}{10}=\frac{6x_0 - 8y_0+18}{10}=\frac{3x_0 - 4y_0+9}{5}$
For each vertex of the given triangle, use the above formulas to find its reflection and then plot the new points to get the reflected figure.
Since no specific coordinates of the vertices of the triangle are given in text form, we can't give numerical answers. But the general method to find the reflected points of a figure over the line $y=\frac{1}{3}x + 1$ is as above.