Math Problem Statement

Solve part 2

The point D is the point on AB such that CD is perpendicular to AB. Calculate the x-coordinate of D.

Solution

Let's break down and solve part (ii) of the problem.

We are given:

  • A(2,1)A(-2, -1)
  • B(4,6)B(4, 6)
  • C(6,3)C(6, -3)

Problem:

The point DD lies on line ABAB such that line CDCD is perpendicular to line ABAB. We are asked to calculate the x-coordinate of DD.

Steps:

  1. Find the slope of ABAB: The slope mABm_{AB} between two points A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) is given by: mAB=y2y1x2x1m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} Substituting the coordinates of A(2,1)A(-2, -1) and B(4,6)B(4, 6): mAB=6(1)4(2)=76m_{AB} = \frac{6 - (-1)}{4 - (-2)} = \frac{7}{6}

  2. Find the slope of CDCD: Since CDCD is perpendicular to ABAB, the slopes of the two lines satisfy: mABmCD=1m_{AB} \cdot m_{CD} = -1 Hence, the slope of CDCD, mCDm_{CD}, is: mCD=1mAB=67m_{CD} = -\frac{1}{m_{AB}} = -\frac{6}{7}

  3. Equation of line ABAB: The equation of a line passing through two points can be written as: yy1=m(xx1)y - y_1 = m(x - x_1) Using the slope mAB=76m_{AB} = \frac{7}{6} and point A(2,1)A(-2, -1): y+1=76(x+2)y + 1 = \frac{7}{6}(x + 2) Simplifying: y=76x+731=76x+43y = \frac{7}{6}x + \frac{7}{3} - 1 = \frac{7}{6}x + \frac{4}{3} So, the equation of line ABAB is: y=76x+43y = \frac{7}{6}x + \frac{4}{3}

  4. Equation of line CDCD: The line CDCD passes through C(6,3)C(6, -3) and has slope mCD=67m_{CD} = -\frac{6}{7}, so its equation is: y(3)=67(x6)y - (-3) = -\frac{6}{7}(x - 6) Simplifying: y+3=67x+367y + 3 = -\frac{6}{7}x + \frac{36}{7} y=67x+3673=67x157y = -\frac{6}{7}x + \frac{36}{7} - 3 = -\frac{6}{7}x - \frac{15}{7} So, the equation of line CDCD is: y=67x157y = -\frac{6}{7}x - \frac{15}{7}

  5. Intersection of lines ABAB and CDCD: To find the coordinates of DD, we need to solve the system of equations of lines ABAB and CDCD.

    Set the equations equal to each other: 76x+43=67x157\frac{7}{6}x + \frac{4}{3} = -\frac{6}{7}x - \frac{15}{7} Let's multiply through by 42 to eliminate the denominators: 42(76x+43)=42(67x157)42\left( \frac{7}{6}x + \frac{4}{3} \right) = 42\left( -\frac{6}{7}x - \frac{15}{7} \right) Simplifying both sides: 49x+56=36x9049x + 56 = -36x - 90 Combine like terms: 49x+36x=905649x + 36x = -90 - 56 85x=14685x = -146 Solving for xx: x=14685=146851.72x = \frac{-146}{85} = -\frac{146}{85} \approx -1.72

Thus, the x-coordinate of point DD is approximately 1.72-1.72.

Would you like further clarification or additional details on any steps? Here are some related questions:

  1. How do we verify if a triangle is isosceles using the distance formula?
  2. Can you explain the process for finding the y-coordinate of DD?
  3. How do perpendicular lines relate in terms of slopes?
  4. What are alternative methods for finding the intersection of two lines?
  5. How can we determine the area of triangle ABCABC using its vertices?

Tip: When solving for the intersection of two lines, always check for computational accuracy by verifying the solution in the original equations.

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Math Problem Analysis

Mathematical Concepts

Analytic Geometry
Perpendicular Slopes
Intersection of Lines

Formulas

Slope of a line: m = (y2 - y1) / (x2 - x1)
Perpendicular slopes: m1 * m2 = -1
Equation of a line: y - y1 = m(x - x1)

Theorems

Perpendicular Line Theorem
Slope-Intercept Theorem

Suitable Grade Level

Grades 9-12