Math Problem Statement

The line y=-4x+32 crosses the y and x axes at points A and B. A line perpendicular to this, that passes through point B, crosses the y-axis at point C. Determine the area of triangle ABC.

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Solution

To solve this problem, let's follow these steps:

  1. Find the coordinates of points A and B where the line y=4x+32y = -4x + 32 crosses the y-axis and x-axis, respectively.
  2. Determine the equation of the perpendicular line that passes through point B and find the coordinates of point C where this line crosses the y-axis.
  3. Calculate the area of triangle ABC using the coordinates of points A, B, and C.

Step 1: Find Points A and B

  • Point A is where the line crosses the y-axis. This occurs when x=0x = 0.

    Substituting x=0x = 0 in the line equation: y=4(0)+32=32y = -4(0) + 32 = 32 So, A=(0,32)A = (0, 32).

  • Point B is where the line crosses the x-axis. This occurs when y=0y = 0.

    Setting y=0y = 0 in the line equation: 0=4x+320 = -4x + 32 Solving for xx: 4x=32    x=84x = 32 \implies x = 8 So, B=(8,0)B = (8, 0).

Step 2: Find the Equation of the Perpendicular Line and Point C

  • The slope of the line y=4x+32y = -4x + 32 is 4-4. The slope of a line perpendicular to this would be the negative reciprocal, which is 14\frac{1}{4}.

  • The equation of the perpendicular line passing through point B=(8,0)B = (8, 0) is of the form: yy1=m(xx1)y - y_1 = m(x - x_1) where (x1,y1)=(8,0)(x_1, y_1) = (8, 0) and m=14m = \frac{1}{4}.

    Substituting these values: y0=14(x8)y - 0 = \frac{1}{4}(x - 8) Simplifying, we get: y=14x2y = \frac{1}{4}x - 2

  • Point C is where this line crosses the y-axis, which occurs when x=0x = 0.

    Substituting x=0x = 0: y=14(0)2=2y = \frac{1}{4}(0) - 2 = -2 So, C=(0,2)C = (0, -2).

Step 3: Calculate the Area of Triangle ABC

We can use the formula for the area of a triangle given by three vertices (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), and (x3,y3)(x_3, y_3):

Area=12x1(y2y3)+x2(y3y1)+x3(y1y2)\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|

Substitute A=(0,32)A = (0, 32), B=(8,0)B = (8, 0), and C=(0,2)C = (0, -2):

Area=120(0(2))+8((2)32)+0(320)\text{Area} = \frac{1}{2} \left| 0(0 - (-2)) + 8((-2) - 32) + 0(32 - 0) \right|

Simplify the expression:

Area=120+8(34)+0=12272\text{Area} = \frac{1}{2} \left| 0 + 8(-34) + 0 \right| = \frac{1}{2} \left| -272 \right|

Area=12×272=136\text{Area} = \frac{1}{2} \times 272 = 136

So, the area of triangle ABC is 136 square units.


Would you like any more details or have any questions about the solution?

Here are some related questions you might find interesting:

  1. What is the equation of a line perpendicular to y=4x+32y = -4x + 32 that passes through a different point, such as (2, 5)?
  2. How can you determine if two lines are perpendicular by looking at their slopes?
  3. What is the geometric significance of the slopes of perpendicular lines being negative reciprocals of each other?
  4. How can you find the area of a triangle if the coordinates of its vertices are given in 3D space?
  5. How does the formula for the area of a triangle change if the triangle is not on a Cartesian plane but on a polar coordinate system?

Tip: When working with perpendicular lines, always remember that their slopes are negative reciprocals of each other. This property is very useful in coordinate geometry.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Slope of a Line
Perpendicular Lines
Area of a Triangle

Formulas

Area of a Triangle formula

Theorems

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Suitable Grade Level

High School