Answer:
[tex]y=-\frac{1}{2}x+2[/tex]
Step-by-step explanation:
So we want to find the equation of a line perpendicular to y=2x+5 and passes through the point (-6,5).
First, let's determine the slope of our equation. Remember that the slopes of perpendicular lines are negative reciprocals. In other words:
[tex]m_1\cdot m_2=-1[/tex]
To find our slope, let's substitute 2 (the slope of y=2x+5) for m₁ and solve for m₂. So:
[tex]2\cdot m_2=-1[/tex]
Divide both sides by 2:
[tex]m_2=-\frac{1}{2}[/tex]
Therefore, the slope of our new line is -1/2.
Now, we can use the point-slope form to find the equation of our line. The point-slope form is:
[tex]y-y_1=m(x-x_1)[/tex]
Where m is the slope and (x₁, y₂) is a point.
So, let's substitute -1/2 for m and (-6,5) for (x₁, y₂), respectively. So:
[tex]y-5=-\frac{1}{2}(x-(-6))[/tex]
Simplify:
[tex]y-5=-\frac{1}{2}(x+6)[/tex]
Distribute:
[tex]y-5=-\frac{1}{2}x-3[/tex]
Add 5 to both sides:
[tex]y=-\frac{1}{2}x+2[/tex]
So, our equation is:
[tex]y=-\frac{1}{2}x+2[/tex]
