Question 332067

We can see that the equation {{{y=-x+3}}} has a slope {{{m=-1}}} and a y-intercept {{{b=3}}}.



Now to find the slope of the perpendicular line, simply flip the slope {{{m=-1}}} to get {{{m=-1/1}}}. Now change the sign to get {{{m=1}}}. So the perpendicular slope is {{{m=1}}}.



Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope {{{m=1}}} and the coordinates of the given point *[Tex \LARGE \left\(-1,2\right\)].



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



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



{{{y-2=1(x+1)}}} Rewrite {{{x--1}}} as {{{x+1}}}



{{{y-2=1x+1(1)}}} Distribute



{{{y-2=1x+1}}} Multiply



{{{y=1x+1+2}}} Add 2 to both sides. 



{{{y=1x+3}}} Combine like terms. 



{{{y=x+3}}} Simplify. 



So the equation of the line perpendicular to {{{y=-x+3}}} that goes through the point *[Tex \LARGE \left\(-1,2\right\)] is {{{y=x+3}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,-x+3,1x+3)
circle(-1,2,0.08),
circle(-1,2,0.10),
circle(-1,2,0.12))}}}


Graph of the original equation {{{y=-x+3}}} (red) and the perpendicular line {{{y=x+3}}} (green) through the point *[Tex \LARGE \left\(-1,2\right\)].