Question 171552


{{{-5x-2y=27}}} Start with the given equation.



{{{-2y=27+5x}}} Add {{{5x}}} to both sides.



{{{-2y=5x+27}}} Rearrange the terms.



{{{y=(5x+27)/(-2)}}} Divide both sides by {{{-2}}} to isolate y.



{{{y=((5)/(-2))x+(27)/(-2)}}} Break up the fraction.



{{{y=-(5/2)x-27/2}}} Reduce.



We can see that the equation {{{y=-(5/2)x-27/2}}} has a slope {{{m=-5/2}}} and a y-intercept {{{b=-27/2}}}.



Now to find the slope of the perpendicular line, simply flip the slope {{{m=-5/2}}} to get {{{m=-2/5}}}. Now change the sign to get {{{m=2/5}}}. So the perpendicular slope is {{{m=2/5}}}.



Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope {{{m=-5/2}}} and the coordinates of the given point *[Tex \LARGE \left\(-5,-2\right\)].



{{{y-y[1]=m(x-x[1])}}} Start with the point slope formula



{{{y--2=(2/5)(x--5)}}} Plug in {{{m=2/5}}}, {{{x[1]=-5}}}, and {{{y[1]=-2}}}



{{{y--2=(2/5)(x+5)}}} Rewrite {{{x--5}}} as {{{x+5}}}



{{{y+2=(2/5)(x+5)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=(2/5)x+(2/5)(5)}}} Distribute



{{{y+2=(2/5)x+2}}} Multiply



{{{y=(2/5)x+2-2}}} Subtract 2 from both sides. 



{{{y=(2/5)x+0}}} Combine like terms. 



{{{y=(2/5)x}}} Remove the trailing zero



So the equation of the line perpendicular to {{{-5x-2y=27}}} that goes through the point *[Tex \LARGE \left\(-5,-2\right\)] is {{{y=(2/5)x}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,-(5/2)x-27/2,(2/5)x)
circle(-5,-2,0.08),
circle(-5,-2,0.10),
circle(-5,-2,0.12))}}}Graph of the original equation {{{y=-(5/2)x-27/2}}} (red) and the perpendicular line {{{y=(2/5)x}}} (green) through the point *[Tex \LARGE \left\(-5,-2\right\)].