Question

Let line $l_1$ be the graph of $5x + 8y = -9$. Line $l_2$ is perpendicular to line $l_1$ and passes through the point $(10,10)$. If line $l_2$ is the graph of the equation $y=mx +b$, then find $m+b$.

Answer 1

Answer:

m + c = - 4.4

Step-by-step explanation:

Line 1 is given by the equation 5x + 8y = -9, ⇒ $y = - \frac{5}{8} x - \frac{9}{5}$ ........(1) {In slope-intercept form}

If line 2 is perpendicular to the line 1 then the equation of line 2 will be  

$y = \frac{8}{5} x + c$  ........ (2), where c is a constant. {Since the product of the slopes of two perpendicular straight line is -1}

Now, the line 2 passes through the point (10,10).

So, from equation (2), we get,

$c = y - \frac{8}{5} x = 10- \frac{8 \times 10}{5} = - 6$

Therefore, the equation of line 2 will be $y = \frac{8}{5} x - 6$

This equation is in y = mx + c form where $m = \frac{8}{5} = 1.6$ and c = - 6

Hence, m + c = 1.6 - 6 = - 4.4 (Answer)