Perpendicular lines

How do you solve this problem?

Are the lines perpendicular?

7y =9x+21

-3y=x-24

They are not perpendicular because:

Line 1:

$7y =9x+21\\ y =\frac98 x+3 \qquad \text{ The slope is :} \quad m_1 = \frac98$

Line 2:

$-3y=x-24\\ y =-\frac{1}{3} x+8 \qquad \text{ The slope is :} \quad m_2 = -\frac{1}{3}$

perpendicular: $\boxed{~ m_1\cdot m_2 = -1 ~ }$

$\begin{array}{lcl} m_1\cdot m_2 &=& \frac98 \cdot (-\frac{1}{3})\\ m_1\cdot m_2 &=& -\frac{9}{8 \cdot3 }\\ m_1\cdot m_2 &=& -\frac{9}{24}\\ m_1\cdot m_2 &=& -\frac{3}{8}\\ m_1\cdot m_2 &=& -0.375 \qquad -0.375\ne -1 \quad \Rightarrow \quad \text{not perpendicular}\\ \end{array}$

Here are the solutions for your equations. You have to graph them to decide if the lines are perpendicular. I myself think they will criss cross. This is how you plot them:

ContourPlot[{7 y = 21 + 9x, -3y = -24 + x}, {x, 71/34, 139/34}, {y, 203/34, 271/34}]

Solve the following system:
{7 y = 9 x+21 |     (equation 1)
-3 y = x-24 |     (equation 2)
Express the system in standard form:
{-(9 x)+7 y = 21 |     (equation 1)
-x-3 y = -24 |     (equation 2)
Subtract 1/9 × (equation 1) from equation 2:
{-(9 x)+7 y = 21 |     (equation 1)
0 x-(34 y)/9 = (-79)/3 |     (equation 2)
Multiply equation 2 by -9:
{-(9 x)+7 y = 21 |     (equation 1)
0 x+34 y = 237 |     (equation 2)
Divide equation 2 by 34:
{-(9 x)+7 y = 21 |     (equation 1)
0 x+y = 237/34 |     (equation 2)
Subtract 7 × (equation 2) from equation 1:
{-(9 x)+0 y = (-945)/34 |     (equation 1)
0 x+y = 237/34 |     (equation 2)
Divide equation 1 by -9:
{x+0 y = 105/34 |     (equation 1)
0 x+y = 237/34 |     (equation 2)
Collect results:
Answer: | 
| {x = 105/34
y = 237/34