Calculus: Early Transcendentals 8th Edition

The product of the slopes of the two lines is $-1$, so these two lines are perpendicular. The point of intersection is $(2,5)$
The slope-intercept form of the equation of a line is $~~y=mx+b~~$ where $m$ is the slope and $b$ is the y-intercept. We can find the slope of the line $3x-5y+19=0$: $3x-5y+19=0$ $5y = 3x+19$ $y = \frac{3}{5}x+\frac{19}{5}$ The slope of the line is $\frac{3}{5}$ We can find the slope of the line $10x+6y-50 = 0$: $10x+6y-50 = 0$ $6y = -10x+50$ $y = -\frac{5}{3}x+\frac{25}{3}$ The slope of the line is $-\frac{5}{3}$ We can find the product of the slopes of the two lines: $(\frac{3}{5})(-\frac{5}{3}) = -1$ If the product of the slopes of two lines is $-1$, then the lines are perpendicular. Therefore, these two lines are perpendicular. We can find the x-coordinate of the point of intersection: $\frac{3}{5}x+\frac{19}{5} = -\frac{5}{3}x+\frac{25}{3}$ $9x+57 = -25x+125$ $34x = 68$ $x = \frac{68}{34}$ $x = 2$ We can find the y-coordinate of the point of intersection: $y = \frac{3}{5}x+\frac{19}{5}$ $y = (\frac{3}{5})(2)+\frac{19}{5}$ $y = \frac{6}{5}+\frac{19}{5}$ $y = \frac{25}{5}$ $y = 5$ The point of intersection is $(2,5)$