+817 Line $\ell_1$ represents the graph of $3x + 2y = -14 - 8x + 5y$. Line $\ell_2$ passes through the point $(5,-6)$, and is perpendicular to line $\ell_1$. Write the equation of line $\ell_2$ in the form $y=mx +b$.
+30 First, we can put \(\ell_1\) into slope intercept form.
3x + 2y = -14 - 8x + 5y
3 y - 11 x = 14
y = 11x/3 + 14/3
Now, if a line is perpendicular to another, their slopes are negative reciprocals of each other. Since \(\ell_1\) has a slope of 11/3 and is perpendicular to \(\ell_2\), \(\ell_2\) will have a slope of -3/11. Plugging this m-value into our slope intercept form y = mx + b yields y = -3x/11 + b. Because (5, -6) is a point on \(\ell_2\), we can plug the x and y-values into the equation to solve for b.
y = -3x/11 + b
(-6) = -3(5)/11 + b
-6 = -15/11 + b
66 = 15 - 11b
11b = -51
b = -51/11
Plug the value of b back into our equation to get the final equation of \(\ell_2\): \(y = -\frac{3}{11}x - \frac{51}{11}\)
