Find the point on the line y = 3x + 5 that is closest to the origin
calculus?
algebra?
we can do either
calculus
i can only do calculus
I prefer the calculus way
First step find the distance from (x,3x+5) to (0,0)
You know the distance formula?
yeah its the distance= sqrt(X2-X1)^2(Y2-Y1)^2)
there should be a + between the x and y my bad
ok so input the points from our question we are finding the distance from (x,y) to (0,0) where it was given in your question y was 3x+5
so its distance= sqrt(X2-X)^2+(3x+5-y)^2)
You need to input the points (x,3x+5) and (0,0)
so sqrt(x^2-(3x+5)^2)
+?
yeah sorry i hate typing equations
\[d^2=x^2+(3x+5)^2 \\ \text{ let } f(x)=x^2+(3x+5)^2 \] find f' then find the critical numbers
so f'= 2x +6(3x+5) then i would set that to 0 to find the critical points right?
yes
so the critical point is X= -20/30
you can reduce that
right to -2/3
and are you sure that is right?
Did you solve 20x+30=0 for x correctly?
Is not that if ax+b=0 then ax=-b then x=-b/a
duh i need to pay attention its -3/2
ok so you find the point by replacing x in y=3x+5 to get y
so y=3(-3/2)+5=?
y=1/2
so you have the point (-3/2,1/2) on y=3x+5 is closest to (0,0) we could have use algebra... y=3x+5 the perpendicular line will have slope -1/3 we want this perpendicular line to go through (0,0) so we have the perpendicular line going through (0,0) is y=-1/3 *x so find the intersection of y=3x+5 and y=-x/3 and we can find the point on y=3x+5 which has the shortest distance from (0,0) so 3x+5=-x/3 multiply 3 on both sides 9x+15=-x 10x+15=0 10x=-15 x=-15/10=-3/2 so y=3x+5=3(-3/2)+5=-9/2+5=-9/2+10/2=1/2 so the point again is (-3/2,1/2)
thanks for all you help I'm getting most of the problems but everyone once and i while i get tripped in easy algebra thanks again and nice profile pic
lol thanks
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