Math Problem Statement

The line y = -4x + 32 crosses the y and x axes at the 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.

Solution

Let's solve the problem step by step based on the image:

Given:

  1. The equation of the line is y=4x+32y = -4x + 32, which crosses the yy-axis at point AA and the xx-axis at point BB.
  2. A perpendicular line to this passes through point BB and crosses the yy-axis at point CC.
  3. We are asked to find the area of triangle ABCABC.

Step 1: Find coordinates of point AA (where the line crosses the yy-axis)

When the line crosses the yy-axis, x=0x = 0.

Substitute x=0x = 0 into the line equation: y=4(0)+32=32y = -4(0) + 32 = 32 Thus, point AA is (0,32)(0, 32).

Step 2: Find coordinates of point BB (where the line crosses the xx-axis)

When the line crosses the xx-axis, y=0y = 0.

Set y=0y = 0 in the equation y=4x+32y = -4x + 32 and solve for xx: 0=4x+320 = -4x + 32 4x=32    x=84x = 32 \implies x = 8 Thus, point BB is (8,0)(8, 0).

Step 3: Find the slope of the perpendicular line and coordinates of point CC

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

This perpendicular line passes through point B(8,0)B(8, 0), so its equation is of the form: yy1=m(xx1)y - y_1 = m(x - x_1) Substituting m=14m = \frac{1}{4}, (x1,y1)=(8,0)(x_1, y_1) = (8, 0): y0=14(x8)y - 0 = \frac{1}{4}(x - 8) y=14x2y = \frac{1}{4}x - 2 To find point CC, set x=0x = 0 (since point CC is on the yy-axis): y=14(0)2=2y = \frac{1}{4}(0) - 2 = -2 Thus, point CC is (0,2)(0, -2).

Step 4: Area of triangle ABCABC

The formula for the area of a triangle given its vertices (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), and (x3,y3)(x_3, y_3) is: 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 the coordinates 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| Area=128(34)=12×272=136\text{Area} = \frac{1}{2} \left| 8(-34) \right| = \frac{1}{2} \times 272 = 136

Thus, the area of triangle ABCABC is 136\boxed{136} square units.

Would you like further clarifications on any step?

5 Related Questions:

  1. How do you find the area of a triangle using the determinant method?
  2. What is the relationship between slopes of perpendicular lines?
  3. How can we derive the equation of a line passing through two points?
  4. How does changing the slope affect the position of the intercepts?
  5. How do we calculate the distance between two points on a coordinate plane?

Tip:

Always double-check the signs of coordinates when calculating areas to avoid errors with negative results!

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Math Problem Analysis

Mathematical Concepts

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

Formulas

Slope of perpendicular lines: m1 * m2 = -1
Area of a triangle: (1/2) * | x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |
Equation of a line: y - y1 = m(x - x1)

Theorems

Slope-intercept form
Properties of perpendicular lines
Formula for area of triangle using vertices

Suitable Grade Level

Grades 10-12