# Find the point on the line y = 4x + 3 that is closest to the origin?

Find the point on the line y = 4x + 3 that is closest to the origin?

• 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(xβΒ² + yβΒ²) = β[(-12/17)Β² + (3/17)Β²] = (3β17)/17 β 0.728

(xβ; yβ) = (-12/17; 3/17)

Methodically,

• y = mx + b where b is the y-intercept (where x = 0)

so 3 is the y-intercept

that is close ( 0, 3 )

Let’s check the x-intercept (where y = 0 )

4x + 3 = 0

4x = – 3

x = – 0.75 that is closer than above, and y-value 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 Ζα΄ΚΡher than * above, new result is point (-0.5, 1)

Let’s try – 0.25 for x

then y = 2 closer still ( -0.25, 2) Ζα΄ΚΡher 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 Ζα΄ΚΡher

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)