Find the point on the line y = 4x + 3 that is closest to the origin?
SOMEONE ASKED π
Find the point on the line y = 4x + 3 that is closest to the origin?
HERE THE ANSWERS π

Hello,
Every point M(x; y) of the line fits the equation:
y = 4x + 3
The distance from such a point M(x; y) to the origin would be:
distance = β(xΒ² + yΒ²)
Since y=4x+3, we would get:
distance = β[xΒ² + (4x + 3)Β²] = β(17xΒ² + 24x + 9)
Normally, to find the minimal value of this distance, we would then need to differentiate this expression versus x and find the value xβ which would make this derivative nil.
However, if the distance is minimal when x=xβ, it stands to reason that the square of that minimal distance when x=xxβ would also be minimal versus the square of all other x. So:
(distance)Β² = 17xΒ² + 24x + 9
[(distance)Β²]’ = 34x + 24
This will be minimal for:
34x + 24 = 0
x = 24/34 = 12/17
So xβ = 12/17
And yβ = 4xβ + 3 = 48/17 + 51/17 = 3/17
And the minimal distance will be:
β(xβΒ² + yβΒ²) = β[(12/17)Β² + (3/17)Β²] = (3β17)/17 β 0.728
(xβ; yβ) = (12/17; 3/17)
Methodically,
Dragon.Jade π

y = mx + b where b is the yintercept (where x = 0)
so 3 is the yintercept
that is close ( 0, 3 )
Let’s check the xintercept (where y = 0 )
4x + 3 = 0
4x = – 3
x = – 0.75 that is closer than above, and yvalue is zero, point (0.75, 0) *
that looks likely to be the one closest to origin
because if we go to x = 1
y = 1 getting further away from (0, 0)
but to be sure let’s try x = 0.5
and y = 5 there, too far away, above origin
but let’s try x = – 0.5
then y = 1 Ζα΄ΚΡher than * above, new result is point (0.5, 1)
Let’s try – 0.25 for x
then y = 2 closer still ( 0.25, 2) Ζα΄ΚΡher away, both values need to be closest to (0,0)
Let’s try x = – 0.6
then y = 0.6 giving ( 0.6, 0.6) closer than * above
since we got closer, we’ll try – 0.55 for x
then y = (0.55)(4) + 3 = 0.8 getting further away
to be sure we’ll try x = – 0.59
then y = 0.64 (0.59, 0.64)
so the bear likes (0.6, 0.6) best, BECAUSE a lower or higher value for x is Ζα΄ΚΡher
Hmmmm, if Dragon head is closer, I should have tried lower value for y, which Dragon says is 3/17
giving 3 = 68x + 51, 68x = – 48, x = – 48/68, = – 12/17 minimizing closest diagonal, a bit fussier than I thought wanted, but by my method 0.70588 seemed further away for x, my diagonal is 0.84853 and Dragon’s is closer, yes.
Source(s): Needs careful checking, bear is not a math head. 
(1,1)