Given the equation of a line, we can use the slope to find the slope of a perpendicular line. If we call m the slope, then the slope of a perpendicular line is:
[tex]Slope\text{ }perpendicular\text{ }line=-\frac{1}{m}[/tex]
Now, we are given the equation of the line:
[tex]\frac{6x+2}{4}-y=1-2y[/tex]
To find the slope, we need to rewrite this in slope-intercept form. To do this, we need to solve for y:
[tex]\begin{gathered} \frac{6x+2}{4}=1-2y+y \\ . \\ \frac{3}{2}x+\frac{1}{2}=1-y \\ . \\ -(\frac{3}{2}x+\frac{1}{2}-1)=y \\ . \\ y=-\frac{3}{2}x+\frac{1}{2} \end{gathered}[/tex]
Then, the slope of this line is -3/2. The slope of a line perpendicular to it is the inverse of the reciprocal:
[tex]Slope\text{ }perpendicular=-\frac{1}{(-\frac{3}{2})}=\frac{2}{3}[/tex]
Now, we know that the slope of the line perpendicular is 2/3, and it must pass through the point (5, -10). We can use the slope- point fomr of a line.
The point-slope form of a line, given slope m and a point P is:
[tex]\begin{gathered} P=(x_P,y_P) \\ . \\ y=m(x-x_P)+y_P \end{gathered}[/tex]
IN the perpendicular line, m = 2/3 and P = (5, -10):
[tex]y=\frac{2}{3}(x-5)-10[/tex]
And solve:
[tex]\begin{gathered} y=\frac{2}{3}x-\frac{2}{3}\cdot5-10 \\ . \\ y=\frac{2}{3}x-\frac{10}{3}-10 \\ . \\ y=\frac{2}{3}x-\frac{40}{3} \end{gathered}[/tex]
The answer is:
[tex]y=\frac{2}{3}x-\frac{40}{3}[/tex]
