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Determine the point on the line y = 4x + 3 that is closest to the origin.

Question:

Determine the point on the line {eq}y = 4x + 3 {/eq} that is closest to the origin.

Distance between the two Points:

First we have to understand the distance between the two points to solve this problem:

Let {eq}(x_1, y_1) {/eq} and {eq}(x_2, y_2) {/eq} be two point on the line, then distance between the two point is given by: {eq}D = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 } {/eq}

Answer and Explanation:


Given:

{eq}y = 4 x + 3 \text{ and point } (0, 0) {/eq}

Let {eq}P(0, 0) {/eq} and {eq}Q {/eq} be a point on the line {eq}y = 4 x + 3 {/eq}.

Assume the x-coordinate of the point Q is t, then the y-coordinate is:

{eq}y = 4 t + 3 {/eq}

Now, we have point Q in the terms of t : {eq}Q(t, 4 t + 3) {/eq}


We will compute the distance between the P and Q.

As we know the distance formula is:

$$\begin{align*} \displaystyle D &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 } &\text{(Where } (x_1, y_1) = (0, 0) \text{ and } (x_2, y_2) = \left (t, 4 t + 3 \right ) \text{)}\\ &= \sqrt{(0 - t)^2 + (0 - (4 t + 3))^2 } &\text{(Plugging in the points)}\\ &= \sqrt{ t^2 + 16 t^2 + 9 + 24 t}\\ D^2 &= 17 t^2 + 24 t + 9 &\text{(Taking square both sides )}\\ &\text{ If we minimize the } D^2 \text{ that means we are minimize } D \\ \frac{d}{dt} [D^2] &= \frac{d}{dx} [ 17 t^2 + 24 t + 9 ] &\text{(Taking derivative with respect to t)}\\ &= 17 \times 2 t + 24 \times 1 + 0 &\text{(Differentiating using the formula } \frac{d}{dx} [(ax + b)^n] = a (ax + b)^{n - 1} \text{)}\\ &= 34 t + 24 \\ \frac{d}{dt} [D^2] &= 0 &\text{(For getting the critical point because the critical point will give minimum } D^2 \text{)}\\ 34 t + 24 &= 0\\ 34 t &= - 24 \\ t &= - \frac{24}{34} &\text{(Dividing both sides by 34)}\\ t &= - \frac{12}{17} \end{align*} $$

Plug the value of t in the point Q

{eq}\displaystyle Q \left (\frac{12}{17}, 4 \times \frac{12}{17} + 3 \right ) = \left ( \frac{12}{17}, \ \frac{99}{17} \right ) {/eq}

Thus the closest to origin is {eq}\displaystyle \left ( \frac{12}{17}, \ \frac{99}{17} \right ) {/eq}.


Learn more about this topic:

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How to Find the Distance Between a Point & a Line

from FTCE Mathematics 6-12 (026): Practice & Study Guide

Chapter 30 / Lesson 5
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