Question 345143


{{{5y+3x=7}}} Start with the given equation.



{{{5y=7-3x}}} Subtract {{{3x}}} from both sides.



{{{5y=-3x+7}}} Rearrange the terms.



{{{y=(-3x+7)/(5)}}} Divide both sides by {{{5}}} to isolate y.



{{{y=((-3)/(5))x+(7)/(5)}}} Break up the fraction.



{{{y=-(3/5)x+7/5}}} Reduce.



We can see that the equation {{{y=-(3/5)x+7/5}}} has a slope {{{m=-3/5}}} and a y-intercept {{{b=7/5}}}.



Now to find the slope of the perpendicular line, simply flip the slope {{{m=-3/5}}} to get {{{m=-5/3}}}. Now change the sign to get {{{m=5/3}}}. So the perpendicular slope is {{{m=5/3}}}.



Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope {{{m=5/3}}} and the coordinates of the given point *[Tex \LARGE \left\(-2,6\right\)].



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



{{{y-6=(5/3)(x--2)}}} Plug in {{{m=5/3}}}, {{{x[1]=-2}}}, and {{{y[1]=6}}}



{{{y-6=(5/3)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y-6=(5/3)x+(5/3)(2)}}} Distribute



{{{y-6=(5/3)x+10/3}}} Multiply



{{{y=(5/3)x+10/3+6}}} Add 6 to both sides. 



{{{y=(5/3)x+28/3}}} Combine like terms. 



So the equation of the line perpendicular to {{{3x+5y=7}}} that goes through the point *[Tex \LARGE \left\(-2,6\right\)] is {{{y=(5/3)x+28/3}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,-(3/5)x+7/5,(5/3)x+28/3)
circle(-2,6,0.08),
circle(-2,6,0.10),
circle(-2,6,0.12))}}}


Graph of the original equation {{{y=-(3/5)x+7/5}}} (red) and the perpendicular line {{{y=(5/3)x+28/3}}} (green) through the point *[Tex \LARGE \left\(-2,6\right\)].