Find the point on the line y= 2x - 3 that is closest to the origin?

Question

i know i'm supposed to use differentiation, but how would that work?

Answer 1

alright so you know two points either you realize or not.

(0, 0) (x,y) where y = 2x + -3
(0,0) (x, 2x+ -3)

Now you have to use the distance formula:

D= √(x² +y²)
D= √(x² + (2x+ -3)²)
D=√ (x² + 4x² + -12x +9)
D=√(5x² + -12x + 9)

Now you have to take the derivative and set it to 0.
D'= 10x+ -12 / 2√(5x² + -12x + 9) = 0
D' = 5x + -6 / √(5x² + -12x + 9) = 0

Since we only need to numerator to = 0 for the whole thing to be 0 we do the following:

5x + -6= 0

Solve for x
5x = 6
x = 5/6

Now we need to find y:

y=2x+ -3
y=2(5/6) + -3
y= -4/3

Answer: (5/6 , -4/3)

Hope that was helpful!

Answer 2

this is how i might address this ... not asserting that's the main appropriate way, or the way you're searching for, yet ... (a million,2,a million) is the conventional vector of airplane a million and (3, a million, 2) is the conventional vector of airplane 2. So taking the go made from those 2 grants a vector perpendicular to *the two* a million and a couple of, so as that would be the direction vector of the intersection. that's (3, a million, -5) We additionally prefer a commencing element for the line, something that's on the two planes, and something easy. How approximately (-5, 0, 9)? So we've a line r = (-5, 0, 9) + t(3, a million, -5) the area from the beginning then is given my Pythagoras,given we merely prefer the minimum we don't prefer the sq. root. (-5+3t)² + (t²) + (9 - 5t)² = 25 - 30t + 9t² + t² + 80 one - 90t + 25t² = 35t² + -120t + a million-6 The minimum of a quadratic happens at -b / 2a So t = one hundred twenty/70 = 12/7 So the closest element is (-5, 0, 9) + 12/7(3, a million, -5) = (0.142, a million.714, 0.40 two) merely examine that element lies on the two planes (0.142) + (2 * a million.714) + (0.40 two) = 4 (3 * 0.142) + a million.714 + (2 * 0.40 two) = 3 So the math is actual in that the element lies on the two planes.

Answer 3

I cheated and plotted your line on graph paper.
I know that the slope of your line [ m =2 ]
I also know the point which is closest to the origin must be on a line that passes through point ( 0,0)
I know the slope of that line must be negative.
I know the line which passes through (0,0) must intersect y=2x-3.
I know that the line that passes through (0,0) and intersects y=2x-3 must have the shortest length from (0,0) to y=2x-3.
By graphing I see that the slope is (m= -1) The point on the line y-2x-3 is ( 1,-1) The length of the line is C = √ x² + y²
C = √ 1² + (-1)²
C= √ 2 = 1.414
I can see that as the slope changes as it rotates about the origin and intersects y=2x-3 the length changes to which we want the minimum value.
I just can not set this up in an equation to be solved. Sorry!

Answer 4

d^2 = (2x-3)^2 + x^2
d^2´= 4(2x-3)+2x = 0 so 10x=12 and x= 6/5 and y = -3/5

Answer 5

The point closest to the origin is y = 0.

So 2x - 3 = 0

Then 2x = 3

So x = 3/2 = 1.5