Question 352953


We can see that the equation {{{y=(3/2)x+5}}} has a slope {{{m=3/2}}} and a y-intercept {{{b=5}}}.



Now to find the slope of the perpendicular line, simply flip the slope {{{m=3/2}}} to get {{{m=2/3}}}. Now change the sign to get {{{m=-2/3}}}. So the perpendicular slope is {{{m=-2/3}}}.



Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope {{{m=-2/3}}} and the coordinates of the given point *[Tex \LARGE \left\(0,0\right\)].



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



{{{y-0=(-2/3)(x-0)}}} Plug in {{{m=-2/3}}}, {{{x[1]=0}}}, and {{{y[1]=0}}}



{{{y-0=(-2/3)x+(-2/3)(0)}}} Distribute



{{{y-0=(-2/3)x+0}}} Multiply



{{{y=(-2/3)x}}} Simplify (ie remove unneeded zeros).



So the equation of the line perpendicular to {{{y=3/2x+5}}} that goes through the point *[Tex \LARGE \left\(0,0\right\)] is {{{y=(-2/3)x}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,(3/2)x+5,(-2/3)x)
circle(0,0,0.08),
circle(0,0,0.10),
circle(0,0,0.12))}}}


Graph of the original equation {{{y=(3/2)x+5}}} (red) and the perpendicular line {{{y=(-2/3)x}}} (green) through the point *[Tex \LARGE \left\(0,0\right\)]. 



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Jim