Show that the lines 3x − 5y + 24 = 0 and 10x + 6y − 56 = 0 are perpendicular. Find both their slopes and find their point of intersection.

Answer 1

3x-5y +24 = 0

y = 3x/5 + 24/5, slope =3/5, the x terms coefficient when the equation is solved for y

10x +6y -56 = 0

y = -5x/3 +28/3 slope =-5/3 = the negative inverse, which is proof the lines are perpendicular

3x/5 + 24/5 = -5x/3 + 28/3

multiply by 15

9x +72 = -25x + 140

34x = 68

x = 68/34 = 2

y= -5(2)/3 + 28/3 = 18/3 =6

intersection point = (2,6)