Draw a line perpendicular to the line that contains the points (1, 8)and (4, 6)and passes through the point (βˆ’2, 8)

Accepted Solution

Answer:[tex]y=\frac{3}{2} x+\frac{20}{3}[/tex]Step-by-step explanation:First, to find the original line, employ the slope formula.[tex]\frac{y2-y1}{x2-x1} \\\\\frac{6-8}{4-1} \\\\\frac{-2}{3}[/tex]The slope of your original line is [tex]-\frac{2}{3}[/tex].Next, plug in your slope and the third point to the point-slope formula.[tex]y-y1=m(x-x1)\\\\y-8=-\frac{2}{3} (x+2)\\\\y-8=-\frac{2}{3} x-\frac{4}{3}\\\\y=-\frac{2}{3} x+\frac{20}{3}[/tex]To find the line which is perpendicular to the line, take the opposite reciprocal of the slope.To find the opposite, flip the sign. [tex]-\frac{2}{3}[/tex] is negative, so it will become [tex]\frac{2}{3}[/tex], which is positive.To find the reciprocal, flip the fraction. [tex]\frac{2}{3}[/tex] would become [tex]\frac{3}{2}[/tex].Your slope for the perpendicular line is [tex]\frac{3}{2}[/tex], so your line is:[tex]y=\frac{3}{2} x+\frac{20}{3}[/tex]