If I'm not mistaken, and I very well could be, this is a calculus problem(?). In order to find the distance without calculus you'd need a point on the given line to use to find the distance in the distance formula. But you don't have a point on the given line, so we can find the shortest distance between the point (1, 4) and the given line using the derivative of the polynomial formed when using the distance formula.

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and we have the x and y for x2 (or x1...it doesn't matter which you choose to fill in):

$d=\sqrt{(1-x)^2+(4-y)^2}$

but what we find is that we have too many unknowns here, namely, the distance, the x coordinate, and the y coordinate. So we can replace the y coordinate with what y is equal to in terms of the linear equation:

$d=\sqrt{(1-x)^2+(4-\frac{1}{3}x-3)^2 }$ and simplify:

$d=\sqrt{(1-x)^2+(7-\frac{1}{3}x)^2 }$

. No we'll expand each binomial by squaring:

$d=\sqrt{(1-2x+x^2)+(49-\frac{14}{3}x+\frac{1}{9}x^2) }$

. Combining like terms gives us

$d=\sqrt{\frac{10}{9}x^2-\frac{20}{3}x+50 }$

The distance between the point (1, 4) and the given line will be at a minimum when the polynomial above is at a minimum. We find the value of x for which the polynomial is at a minimum by finding its derivative, setting the derivative equal to 0, and then solving for x. The derivative of the polynomial is

$\frac{20}{9}x-\frac{20}{3}$

Setting equal to 0 and getting rid of the denominators gives us

20x - 60 = 0

Solving for x gives us

20x = 60 and x = 3.

That's the value of x that gives us the shortest distance between (1, 4) and the line y = 1/3x - 3. Sub into the distance formula that x value to find the distance:

$d=\sqrt{(\frac{10}{9})(3)^2-(\frac{20}{3})(3)+50 }$

which simplifies down, finally, to

x ≈ 6.325 units

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3 years ago