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

m + c = - 4.4

Explanation:

Line 1 is given by the equation 5x + 8y = -9, ⇒
`y = - (5)/(8) x - (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 = (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 - (8)/(5) x = 10- (8 * 10)/(5) = - 6`

Therefore, the equation of line 2 will be
`y = (8)/(5) x - 6`

This equation is in y = mx + c form where
`m = (8)/(5) = 1.6` and c = - 6

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