Answer
The product of the slopes of the two lines is $-1$, so these two lines are perpendicular.
The point of intersection is $(2,5)$
Work Step by Step
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 $(2,5)$
