Math Problem Statement

Find the point in the line y=2x-3 that is closest to origin

Solution

To find the point on the line y=2x3y = 2x - 3 that is closest to the origin (0,0)(0,0), we can follow these steps:

Step 1: Define the point on the line

Let's denote the point on the line by P(x1,y1)P(x_1, y_1). Since the point lies on the line y=2x3y = 2x - 3, we can express y1y_1 as: y1=2x13y_1 = 2x_1 - 3

Step 2: Find the distance from the origin to the point

The distance DD from the origin (0,0)(0,0) to the point P(x1,y1)P(x_1, y_1) is given by the distance formula: D=(x10)2+(y10)2=x12+y12D = \sqrt{(x_1 - 0)^2 + (y_1 - 0)^2} = \sqrt{x_1^2 + y_1^2} Substituting y1=2x13y_1 = 2x_1 - 3 into the distance formula, we get: D=x12+(2x13)2D = \sqrt{x_1^2 + (2x_1 - 3)^2}

Step 3: Minimize the distance

To minimize the distance DD, we can minimize D2D^2 instead, as it will make the calculation easier. So, we minimize: D2=x12+(2x13)2D^2 = x_1^2 + (2x_1 - 3)^2 Expanding the square: D2=x12+(4x1212x1+9)=5x1212x1+9D^2 = x_1^2 + (4x_1^2 - 12x_1 + 9) = 5x_1^2 - 12x_1 + 9

Step 4: Differentiate and find the minimum

To find the minimum, differentiate D2D^2 with respect to x1x_1 and set the derivative equal to zero: d(D2)dx1=10x112=0\frac{d(D^2)}{dx_1} = 10x_1 - 12 = 0 Solve for x1x_1: 10x1=12x1=6510x_1 = 12 \quad \Rightarrow \quad x_1 = \frac{6}{5}

Step 5: Find the corresponding y1y_1

Substitute x1=65x_1 = \frac{6}{5} back into the equation y1=2x13y_1 = 2x_1 - 3: y1=2(65)3=125155=35y_1 = 2\left(\frac{6}{5}\right) - 3 = \frac{12}{5} - \frac{15}{5} = \frac{-3}{5}

Final Answer

The point on the line y=2x3y = 2x - 3 that is closest to the origin is (65,35)\left(\frac{6}{5}, \frac{-3}{5}\right).

Would you like more details or have any further questions?

Related Questions:

  1. How do you derive the distance formula between two points in 2D space?
  2. Can you explain how to find the derivative of a quadratic function?
  3. What is the geometric interpretation of minimizing the distance to a line?
  4. How would you find the point closest to the origin on a different line, say y=x+4y = -x + 4?
  5. What if the line was a curve, how would you find the closest point to the origin?
  6. Can you solve this problem using vector projection?
  7. What is the significance of the slope in determining the closest point?
  8. How do perpendicular distances from a point to a line relate to the shortest distance?

Tip:

When finding the minimum or maximum distance between a point and a curve (line), consider minimizing the square of the distance to avoid dealing with square roots in the differentiation process.

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Math Problem Analysis

Mathematical Concepts

Analytic Geometry
Distance Formula
Minimization

Formulas

Distance formula
Quadratic minimization

Theorems

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Suitable Grade Level

Grades 11-12