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

Question

Solve the point on the line&amp;nbsp;y=2x+3&amp;nbsp;that is closest to the origin.

Neelam Wainwright · Accepted Answer

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Mr Solver · Accepted Answer

Answer:(−43,−53)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&#039;s assume that the point on the line closest to the origin is (x0,y0). The distance between the origin (0,0) and the point (x0,y0) is given by the formula:distance=(x0−0)2+(y0−0)2Since the point (x0,y0) lies on the line y=2x+3, we can substitute y in terms of x:distance=(x0−0)2+((2x0+3)−0)2Simplifying the equation further:distance=x02+(2x0+3)2To minimize the distance, we can differentiate the equation with respect to x0 and set the derivative equal to zero:d(distance)dx0=0Differentiating and solving for x0:ddx0(x02+(2x0+3)2)=0x0x02+(2x0+3)2+2(2x0+3)x02+(2x0+3)2=0x0+4x0+6+4x0+6=09x0=−12x0=−43Now, substituting this value back into the equation for the line y=2x+3 to find y0:y0=2(−43)+3y0=−53Therefore, the point on the line y=2x+3 that is closest to the origin is (−43,−53).

Nick Camelot · Accepted Answer

To solve the problem of finding the point on the line

y = 2 x + 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

(x_{0}, y_{0})

. The distance between the origin

(0, 0)

and the point

(x_{0}, y_{0})

is given by the formula:

distance = \sqrt{(x_{0} - 0)^{2} + (y_{0} - 0)^{2}}

Since the point

(x_{0}, y_{0})

lies on the line

y = 2 x + 3

, we can substitute

y

in terms of

x

:

distance = \sqrt{(x_{0} - 0)^{2} + ((2 x_{0} + 3) - 0)^{2}}

Simplifying the equation further:

distance = \sqrt{x_{0}^{2} + (2 x_{0} + 3)^{2}}

To minimize the distance, we can differentiate the equation with respect to

x_{0}

and set the derivative equal to zero:

\frac{d (distance)}{d x_{0}} = 0

Differentiating and solving for

x_{0}

:

\frac{d}{d x_{0}} (\sqrt{x_{0}^{2} + (2 x_{0} + 3)^{2}}) = 0

\frac{x_{0}}{\sqrt{x_{0}^{2} + (2 x_{0} + 3)^{2}}} + \frac{2 (2 x_{0} + 3)}{\sqrt{x_{0}^{2} + (2 x_{0} + 3)^{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 = 2 x + 3

to find

y_{0}

:

y_{0} = 2 (- \frac{4}{3}) + 3

y_{0} = - \frac{5}{3}

Therefore, the point on the line

y = 2 x + 3

that is closest to the origin is

(- \frac{4}{3}, - \frac{5}{3})

.

Jazz Frenia · Accepted Answer

Step 1: Find the distance between the origin

(0, 0)

and an arbitrary point

(x, y)

on the line

y = 2 x + 3

. This distance can be calculated using the distance formula:

D = \sqrt{(x - 0)^{2} + (y - 0)^{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} + (2 x + 3)^{2}}

Step 3: Simplify the expression under the square root:

D = \sqrt{x^{2} + (4 x^{2} + 12 x + 9)}

Step 4: Expand and combine like terms:

D = \sqrt{5 x^{2} + 12 x + 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{d D}{d x} = \frac{10 x + 12}{2 \sqrt{5 x^{2} + 12 x + 9}} = 0

Simplifying the equation:

10 x + 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 (- \frac{6}{5}) + 3

Simplifying:

y = - \frac{12}{5} + 3 = - \frac{12}{5} + \frac{15}{5} = \frac{3}{5}

Therefore, the point on the line

y = 2 x + 3

that is closest to the origin is

(- \frac{6}{5}, \frac{3}{5})

.

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

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