The vertices of â–ł MQN will fall on lines that are perpendicular to y=- 2x.
Practice makes perfect
To reflect the figure in y=- 2x, we first have to find the perpendicular slope to - 2. Because slopes of perpendicular lines are opposite reciprocals, we know that all perpendicular lines to y=- 2x will have the following format.
y=1/2x+b
Therefore, any point on the image will be on a line that goes through the corresponding point on the preimage and with a slope of 12. Let's show this in a diagram using three different lines, one for each vertex.
To reflect the vertices of â–ł MQN, we must know the equation of all these lines. From the diagram, we can see that the uppermost line intercepts the y-axis at (0,3) to it has the following equation.
y=1/2x+3
To find the remaining two lines, we substitute the coordinates of the vertices, in their respective equation and solve for the y-intercept.
Red Line:& 0=1/2(- 5)+b_1 ⇔ b_1=2.5 [0.8em]
Purple Line:& - 1=1/2(- 1)+b_2 ⇔ b_2=- 0.5
Corresponding vertices on the image and preimage form a segment whose perpendicular bisector is the line of reflection. This means each lines point of intersection with y=- 2x will be the midpoint of these segments. Since we know the equations of each line, we can find the x-coordinate of this midpoint by equating them with y=- 2x.
Lime Line:& - 2x=1/2x+3 ⇔ x=- 1.2 [0.8em]
Red Line:& - 2x=1/2x+2.5 ⇔ x=- 1 [0.8em]
Purple Line:& - 2x=1/2x-0.5 ⇔ x= 0.2
Having found the x-coordinate where the three lines intersect with the line of reflection, we have to solve for the corresponding y-coordinate as well.
Lime Line:& y=1/2(- 1.2)+3 ⇔ y= 2.4 [0.8em]
Red Line:& y=1/2(- 1)+2.5 ⇔ y= 2 [0.8em]
Purple Line:& y=1/2* 0.2-0.5 ⇔ y= - 0.4
When we know each line's intersection point with y=- 2x, we can use the Midpoint Formula to find the coordinates of the image vertices.
Finding M'
We are looking for (x_(M'),y_(M')). By substituting the midpoint and the coordinates for M on the preimage into the midpoint-formula, we can solve for these coordinates.
