1. ## lines perpendicular

hey guys, need help with this .. any suggestions with how i should start with question? im not looking for the answer, i want to learn how to do it..thank u

The lines given by
ax + by = 3 and cx 4y = 5 are perpendicular and intersect at (6, 7). Find a, b, and c.

2. Originally Posted by jvignacio
hey guys, need help with this .. any suggestions with how i should start with question? im not looking for the answer, i want to learn how to do it..thank u

The lines given by
ax + by = 3 and cx 4y = 5 are perpendicular and intersect at (6, 7). Find a, b, and c.
We have the lines $\displaystyle ax+by=3$ and $\displaystyle cx-4y=5$

They are perpendicular and intersect at (6,7)

Recall that the slopes of perpendicular lines are opposite reciprocals of each other.

So we want to find $\displaystyle \frac{\,dy}{\,dx}$ of both lines.

So we see that $\displaystyle \frac{\,dy}{\,dx}=-\frac{a}{b}$ and $\displaystyle \frac{\,dy}{\,dx}=\frac{c}{4}$

Now, let us evaluate the lines at the intersection point:

$\displaystyle 6a+7b=3$-----(1)
$\displaystyle 6c=33$--------(2)

We can now solve for c:

$\displaystyle 6c=33\implies c=\dots$

Now substitute this value into the derivative.

$\displaystyle \frac{\,dy}{\,dx}=\frac{c}{4}$

We see that the negative reciprocal of this slope will give us a slope parallel to the other line. We want these slopes to be equal.

Thus, we must let $\displaystyle -\frac{a}{b}=-\frac{4}{c}$

We must solve this system of equations in order to find a and b.

$\displaystyle \left\{\begin{array}{r}6a+7b=3\\ \frac{a}{b}=\frac{4}{c}\end{array}\right.$

I hope this makes sense!

--Chris