For this case we have the following line equations:[tex]3x + 7y = 15\\7x-3y = 6[/tex]
We manipulate the equations algebraically until we write them in the slope-intersection form,[tex]y = mx + b:[/tex]
Equation 1:
[tex]3x + 7y = 15\\7y = -3x + 15\\y = - \frac {3} {7} x + \frac {15} {7}[/tex]
Equation 2:
[tex]7x-3y = 6\\-3y = -7x + 6\\y = \frac {-7} {- 3} x + \frac {6} {- 3}\\y = \frac {7} {3} x-2[/tex]
By definition we have:
If two lines are parallel then their slopes are equal.
If two lines are perpendicular then the product of their slopes is -1.
It is noted that the slopes are not equal. We check if the product is -1:
[tex]- \frac {3} {7} * \frac {7} {3} = - \frac {3 * 7} {7 * 3} = \frac {-21} {21} = - 1[/tex]
Thus, the lines are perpendicular.
Answer:
The lines are perpendicular
