Question 207797


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



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



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



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



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



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



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



Now to find the slope of the perpendicular line, simply flip the slope {{{m=-7/3}}} to get {{{m=-3/7}}}. Now change the sign to get {{{m=3/7}}}. So the perpendicular slope is {{{m=3/7}}}.



Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope {{{m=-7/3}}} and the coordinates of the given point *[Tex \LARGE \left\(7,-6\right\)].



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



{{{y--6=(3/7)(x-7)}}} Plug in {{{m=3/7}}}, {{{x[1]=7}}}, and {{{y[1]=-6}}}



{{{y+6=(3/7)(x-7)}}} Rewrite {{{y--6}}} as {{{y+6}}}



{{{y+6=(3/7)x+(3/7)(-7)}}} Distribute



{{{y+6=(3/7)x-3}}} Multiply



{{{y=(3/7)x-3-6}}} Subtract 6 from both sides. 



{{{y=(3/7)x-9}}} Combine like terms. 



So the equation of the line perpendicular to {{{7x+3y=5}}} that goes through the point *[Tex \LARGE \left\(7,-6\right\)] is {{{y=(3/7)x-9}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,-(7/3)x+5/3,(3/7)x-9)
circle(7,-6,0.08),
circle(7,-6,0.10),
circle(7,-6,0.12))}}}Graph of the original equation {{{y=-(7/3)x+5/3}}} (red) and the perpendicular line {{{y=(3/7)x-9}}} (green) through the point *[Tex \LARGE \left\(7,-6\right\)].