For this case we have the following line equations:
![3x + 7y = 15\\7x-3y = 6](https://img.qammunity.org/2020/formulas/mathematics/middle-school/j494igqlt2nqore97bn0bja2an9l12n6sg.png)
We manipulate the equations algebraically until we write them in the slope-intersection form,
![y = mx + b:](https://img.qammunity.org/2020/formulas/mathematics/middle-school/butfcsqlxdatcma65u5mcxav62869jyokd.png)
Equation 1:
![3x + 7y = 15\\7y = -3x + 15\\y = - \frac {3} {7} x + \frac {15} {7}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/nt3gtkmct7kmcs9sdp781hsvkrnp8b5xgn.png)
Equation 2:
![7x-3y = 6\\-3y = -7x + 6\\y = \frac {-7} {- 3} x + \frac {6} {- 3}\\y = \frac {7} {3} x-2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/f79ue87c9tht00eaeqg3s0njwd7fx5wbnq.png)
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:
![- \frac {3} {7} * \frac {7} {3} = - \frac {3 * 7} {7 * 3} = \frac {-21} {21} = - 1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/8cfd4rlfcebwr5aole9hr21u1rt5yhh315.png)
Thus, the lines are perpendicular.
Answer:
The lines are perpendicular