Math Problem Statement

The points A(-3, -4) and C(5,4) are the ends of the diagonal of a rhombus ABCD. a Find the equation of the diagonal BD. b Given that the side BC has gradient 5/3, find the coordinates of B and hence of D.

Solution

To solve this problem, let's go step by step.

Part (a): Finding the Equation of the Diagonal BDBD

  1. Find the midpoint of ACAC: Since AA and CC are the ends of a diagonal of the rhombus, the diagonals of a rhombus bisect each other at right angles. The midpoint of ACAC will be the point where both diagonals ACAC and BDBD intersect.

    The midpoint MM of A(3,4)A(-3, -4) and C(5,4)C(5, 4) is calculated as: M=(3+52,4+42)=(22,02)=(1,0)M = \left( \frac{-3 + 5}{2}, \frac{-4 + 4}{2} \right) = \left( \frac{2}{2}, \frac{0}{2} \right) = (1, 0)

  2. Calculate the slope of ACAC: The slope mACm_{AC} between A(3,4)A(-3, -4) and C(5,4)C(5, 4) is: mAC=4(4)5(3)=88=1m_{AC} = \frac{4 - (-4)}{5 - (-3)} = \frac{8}{8} = 1

  3. Determine the slope of BDBD: Since BDBD is perpendicular to ACAC, the slope of BDBD, denoted mBDm_{BD}, is the negative reciprocal of mACm_{AC}: mBD=1m_{BD} = -1

  4. Write the equation of line BDBD: Using the point-slope form of a line yy1=m(xx1)y - y_1 = m(x - x_1) with point M(1,0)M(1, 0) and slope m=1m = -1: y0=1(x1)y - 0 = -1(x - 1) Simplifying, we get: y=x+1y = -x + 1 So, the equation of the diagonal BDBD is: y=x+1y = -x + 1

Part (b): Finding the Coordinates of Points BB and DD

  1. Determine the slope of BCBC: We are given that the slope of BCBC is 53\frac{5}{3}.

  2. Set up the equations for point BB: Let the coordinates of BB be (x,y)(x, y).

    Since M(1,0)M(1, 0) is the midpoint of BCBC, we can use the midpoint formula: 1=x+52and0=y+421 = \frac{x + 5}{2} \quad \text{and} \quad 0 = \frac{y + 4}{2}

    Solving these equations:

    • For xx: 21=x+5x=25=32 \cdot 1 = x + 5 \Rightarrow x = 2 - 5 = -3
    • For yy: 20=y+4y=42 \cdot 0 = y + 4 \Rightarrow y = -4 So, the coordinates of BB are (3,4)(-3, -4).
  3. Find the coordinates of DD: Since MM is also the midpoint of ADAD, and the coordinates of AA are (3,4)(-3, -4), we can apply the midpoint formula again to find DD (we’ll let the coordinates of DD be (x,y)(x', y')): 1=3+x2and0=4+y21 = \frac{-3 + x'}{2} \quad \text{and} \quad 0 = \frac{-4 + y'}{2}

    Solving these:

    • For xx': 21=3+xx=2+3=52 \cdot 1 = -3 + x' \Rightarrow x' = 2 + 3 = 5
    • For yy': 20=4+yy=42 \cdot 0 = -4 + y' \Rightarrow y' = 4

    So, the coordinates of DD are (5,4)(5, 4).

Summary of Results

  1. The equation of the diagonal BDBD is: y=x+1y = -x + 1

  2. The coordinates of BB are (3,4)(-3, -4), and the coordinates of DD are (5,4)(5, 4).

Would you like a detailed explanation of any specific step, or do you have any questions?


Related Questions:

  1. How do you find the midpoint of a line segment in a coordinate plane?
  2. What is the significance of the diagonals in a rhombus and how do they relate to each other?
  3. How do you determine the equation of a line perpendicular to a given line?
  4. What is the process for finding the equation of a line given a slope and a point on the line?
  5. How can you verify if four points form a rhombus on a coordinate plane?

Tip: When solving for coordinates on a rhombus, always remember that the diagonals bisect each other at right angles and each diagonal divides the rhombus into two congruent triangles.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Line Equations
Midpoint Theorem
Perpendicular Lines
Slope
Rhombus Properties

Formulas

Midpoint Formula: M = ((x1 + x2)/2, (y1 + y2)/2)
Slope Formula: m = (y2 - y1) / (x2 - x1)
Equation of a Line: y - y1 = m(x - x1)

Theorems

Diagonals of a Rhombus Bisect Each Other at Right Angles
Perpendicular Lines Have Negative Reciprocal Slopes

Suitable Grade Level

Grades 9-11