Math Problem Statement

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

Solution

To find the point on the line y=2x+3y = 2x + 3 that is closest to the origin, we need to minimize the distance from a point (x,y)(x, y) on the line to the origin (0,0)(0, 0). The distance dd from the origin to a point (x,y)(x, y) is given by:

d=x2+y2d = \sqrt{x^2 + y^2}

Since y=2x+3y = 2x + 3, we can substitute for yy in the distance formula:

d=x2+(2x+3)2d = \sqrt{x^2 + (2x + 3)^2}

To minimize dd, it suffices to minimize d2=x2+(2x+3)2d^2 = x^2 + (2x + 3)^2 because the square root is a monotonically increasing function.

Expand (2x+3)2(2x + 3)^2:

(2x+3)2=4x2+12x+9(2x + 3)^2 = 4x^2 + 12x + 9

Thus, d2d^2 becomes:

d2=x2+4x2+12x+9=5x2+12x+9d^2 = x^2 + 4x^2 + 12x + 9 = 5x^2 + 12x + 9

To find the minimum, take the derivative of d2d^2 with respect to xx and set it to zero:

ddx(5x2+12x+9)=10x+12\frac{d}{dx}(5x^2 + 12x + 9) = 10x + 12

Set the derivative equal to zero:

10x+12=010x + 12 = 0

Solve for xx:

x = -\frac{6}{5}$$ Substitute $$x = -\frac{6}{5}$$ back into the line equation to find $$y$$: $$y = 2\left(-\frac{6}{5}\right) + 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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Math Problem Analysis

Mathematical Concepts

Algebra
Distance Formula
Optimization

Formulas

Distance formula: d = sqrt(x^2 + y^2)
Derivative to find minimum: f'(x) = 0

Theorems

Monotonicity of square root function
Optimization principles

Suitable Grade Level

Grades 10-12