Show that the line 3y = 7 + 5x is perpendicular to the line 5y = 4 - 3x.

Proven Below

Step-by-step explanation:

If we rearrange these equations, we get that 3y = 5x + 7 and that 5y = -3x + 4. Next, to find what y equals on both equations, we divide by 3 on both sides of the first equation, where we would get that y = 5/3x + 7/3, and then divide by 5 on both sides of the second equation, where we would get that y = -3/5x + 4/5. Now, to see if two lines are perpendicular to each other, we have to see if the slopes (The number being multiplied by x. Ex.) In the equation y = 3/4x + 4, the slope is 3/4) are opposite reciprocals (If you multiply them and they are equal to -1. Ex.) 3/4 * -4/3 is equal to -1, so they are opposite reciprocals). In this case, we have -3/5 and 5/3 as slopes, and -3/5 * 5/3 is equal to -1, so they are perpendicular.

The slope of the first line is 5/3

The slope for the second one is -3/5

If you multiply the 2 slope you have to end up with - 1

So 5/3*(-3/5)=-15/15 which is - 1

So yes the lines are perpenducular

y -5 = (-1/4)(x + 8)

step-by-step explanation: