kuCAu

2021-05-17

Solve the point on the line $y=2x+3$ that is closest to the origin.

Neelam Wainwright

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Mr Solver

$\left(-\frac{4}{3},-\frac{5}{3}\right)$
Explanation:
To solve the problem of finding the point on the line $y=2x+3$ that is closest to the origin, we can use the concept of distance between two points.
Let's assume that the point on the line closest to the origin is $\left({x}_{0},{y}_{0}\right)$. The distance between the origin $\left(0,0\right)$ and the point $\left({x}_{0},{y}_{0}\right)$ is given by the formula:
$\text{distance}=\sqrt{\left({x}_{0}-0{\right)}^{2}+\left({y}_{0}-0{\right)}^{2}}$
Since the point $\left({x}_{0},{y}_{0}\right)$ lies on the line $y=2x+3$, we can substitute $y$ in terms of $x$:
$\text{distance}=\sqrt{\left({x}_{0}-0{\right)}^{2}+\left(\left(2{x}_{0}+3\right)-0{\right)}^{2}}$
Simplifying the equation further:
$\text{distance}=\sqrt{{x}_{0}^{2}+\left(2{x}_{0}+3{\right)}^{2}}$
To minimize the distance, we can differentiate the equation with respect to ${x}_{0}$ and set the derivative equal to zero:
$\frac{d\left(\text{distance}\right)}{d{x}_{0}}=0$
Differentiating and solving for ${x}_{0}$:
$\frac{d}{d{x}_{0}}\left(\sqrt{{x}_{0}^{2}+\left(2{x}_{0}+3{\right)}^{2}}\right)=0$
$\frac{{x}_{0}}{\sqrt{{x}_{0}^{2}+\left(2{x}_{0}+3{\right)}^{2}}}+\frac{2\left(2{x}_{0}+3\right)}{\sqrt{{x}_{0}^{2}+\left(2{x}_{0}+3{\right)}^{2}}}=0$
${x}_{0}+4{x}_{0}+6+4{x}_{0}+6=0$
$9{x}_{0}=-12$
${x}_{0}=-\frac{4}{3}$
Now, substituting this value back into the equation for the line $y=2x+3$ to find ${y}_{0}$:
${y}_{0}=2\left(-\frac{4}{3}\right)+3$
${y}_{0}=-\frac{5}{3}$
Therefore, the point on the line $y=2x+3$ that is closest to the origin is $\left(-\frac{4}{3},-\frac{5}{3}\right)$.

Nick Camelot

To solve the problem of finding the point on the line $y=2x+3$ that is closest to the origin, we can use the concept of distance between two points.
Let's assume that the point on the line closest to the origin is $\left({x}_{0},{y}_{0}\right)$. The distance between the origin $\left(0,0\right)$ and the point $\left({x}_{0},{y}_{0}\right)$ is given by the formula:
$\text{distance}=\sqrt{\left({x}_{0}-0{\right)}^{2}+\left({y}_{0}-0{\right)}^{2}}$
Since the point $\left({x}_{0},{y}_{0}\right)$ lies on the line $y=2x+3$, we can substitute $y$ in terms of $x$:
$\text{distance}=\sqrt{\left({x}_{0}-0{\right)}^{2}+\left(\left(2{x}_{0}+3\right)-0{\right)}^{2}}$
Simplifying the equation further:
$\text{distance}=\sqrt{{x}_{0}^{2}+\left(2{x}_{0}+3{\right)}^{2}}$
To minimize the distance, we can differentiate the equation with respect to ${x}_{0}$ and set the derivative equal to zero:
$\frac{d\left(\text{distance}\right)}{d{x}_{0}}=0$
Differentiating and solving for ${x}_{0}$:
$\frac{d}{d{x}_{0}}\left(\sqrt{{x}_{0}^{2}+\left(2{x}_{0}+3{\right)}^{2}}\right)=0$
$\frac{{x}_{0}}{\sqrt{{x}_{0}^{2}+\left(2{x}_{0}+3{\right)}^{2}}}+\frac{2\left(2{x}_{0}+3\right)}{\sqrt{{x}_{0}^{2}+\left(2{x}_{0}+3{\right)}^{2}}}=0$
${x}_{0}+4{x}_{0}+6+4{x}_{0}+6=0$
$9{x}_{0}=-12$
${x}_{0}=-\frac{4}{3}$
Now, substituting this value back into the equation for the line $y=2x+3$ to find ${y}_{0}$:
${y}_{0}=2\left(-\frac{4}{3}\right)+3$
${y}_{0}=-\frac{5}{3}$
Therefore, the point on the line $y=2x+3$ that is closest to the origin is $\left(-\frac{4}{3},-\frac{5}{3}\right)$.

Jazz Frenia

Step 1: Find the distance between the origin $\left(0,0\right)$ and an arbitrary point $\left(x,y\right)$ on the line $y=2x+3$. This distance can be calculated using the distance formula:
$D=\sqrt{\left(x-0{\right)}^{2}+\left(y-0{\right)}^{2}}=\sqrt{{x}^{2}+{y}^{2}}$
Step 2: Substitute the expression for $y$ from the equation of the line into the distance formula:
$D=\sqrt{{x}^{2}+\left(2x+3{\right)}^{2}}$
Step 3: Simplify the expression under the square root:
$D=\sqrt{{x}^{2}+\left(4{x}^{2}+12x+9\right)}$
Step 4: Expand and combine like terms:
$D=\sqrt{5{x}^{2}+12x+9}$
Step 5: To minimize the distance $D$, we need to find the value of $x$ that minimizes the expression under the square root. To find this, we can take the derivative of $D$ with respect to $x$ and set it equal to zero:
$\frac{dD}{dx}=\frac{10x+12}{2\sqrt{5{x}^{2}+12x+9}}=0$
Simplifying the equation:
$10x+12=0$
Step 6: Solve for $x$:
$x=-\frac{12}{10}=-\frac{6}{5}$
Step 7: Substitute the value of $x$ back into the equation of the line to find the corresponding $y$:
$y=2\left(-\frac{6}{5}\right)+3$
Simplifying:
$y=-\frac{12}{5}+3=-\frac{12}{5}+\frac{15}{5}=\frac{3}{5}$
Therefore, the point on the line $y=2x+3$ that is closest to the origin is $\left(-\frac{6}{5},\frac{3}{5}\right)$.

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