Make me a question that requires graphing 3 lines to make a triangle and finding the perimeter of that triangle.
Understand the Problem
The question is asking for the creation of a mathematical problem that involves graphing three lines to form a triangle, and then calculating the perimeter of that triangle. This entails finding the equations of three lines, identifying their intersection points, and then determining the lengths of the triangle's sides to calculate the perimeter.
Answer
The perimeter of the triangle is $\frac{\sqrt{2} + 2\sqrt{5}}{3}$.
Answer for screen readers
The perimeter of the triangle is $\frac{\sqrt{2} + 2\sqrt{5}}{3}$.
Steps to Solve
- Define the Equations of the Lines
Choose three linear equations that will intersect at three points to form a triangle. For example:
- Line 1: $y = 2x + 1$
- Line 2: $y = -x + 4$
- Line 3: $y = \frac{1}{2}x + 2$
- Find the Intersection Points
Calculate the intersection points of the lines by solving the equations in pairs.
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Intersection of Line 1 and Line 2: Set $2x + 1 = -x + 4$. $$ 2x + x = 4 - 1 \ 3x = 3 \ x = 1 $$ Substituting back into Line 1 for $y$: $$ y = 2(1) + 1 = 3 $$ So, Point A is $(1, 3)$.
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Intersection of Line 2 and Line 3: Set $-x + 4 = \frac{1}{2}x + 2$. $$ -x - \frac{1}{2}x = 2 - 4 \ -\frac{3}{2}x = -2 \ x = \frac{4}{3} $$ Substituting back into Line 2 for $y$: $$ y = -\frac{4}{3} + 4 = \frac{8}{3} $$ So, Point B is $\left(\frac{4}{3}, \frac{8}{3}\right)$.
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Intersection of Line 3 and Line 1: Set $\frac{1}{2}x + 2 = 2x + 1$. $$ \frac{1}{2}x - 2x = 1 - 2 \ -\frac{3}{2}x = -1 \ x = \frac{2}{3} $$ Substituting back into Line 3 for $y$: $$ y = \frac{1}{2} \times \frac{2}{3} + 2 = \frac{1}{3} + 2 = \frac{7}{3} $$ So, Point C is $\left(\frac{2}{3}, \frac{7}{3}\right)$.
- Calculate the Lengths of the Triangle Sides
Use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to find the side lengths.
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Length AB: $$ AB = \sqrt{\left(\frac{4}{3} - 1\right)^2 + \left(\frac{8}{3} - 3\right)^2} \ = \sqrt{\left(\frac{1}{3}\right)^2 + \left(-\frac{1}{3}\right)^2} = \sqrt{\frac{1}{9} + \frac{1}{9}} = \sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{3} $$
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Length BC: $$ BC = \sqrt{\left(\frac{2}{3} - \frac{4}{3}\right)^2 + \left(\frac{7}{3} - \frac{8}{3}\right)^2} \ = \sqrt{\left(-\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} $$
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Length CA: $$ CA = \sqrt{(1 - \frac{2}{3})^2 + (3 - \frac{7}{3})^2} \ = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2} = \sqrt{\frac{1}{9} + \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} $$
- Calculate the Perimeter of the Triangle
Add the lengths of the sides to find the perimeter. $$ \text{Perimeter} = AB + BC + CA \ = \frac{\sqrt{2}}{3} + \frac{\sqrt{5}}{3} + \frac{\sqrt{5}}{3} \ = \frac{\sqrt{2} + 2\sqrt{5}}{3} $$
The perimeter of the triangle is $\frac{\sqrt{2} + 2\sqrt{5}}{3}$.
More Information
This problem demonstrates how to find the perimeter of a triangle formed by the intersection of three linear equations in the Cartesian plane. The approach involves determining the equation of each line, their intersections, and applying the distance formula to calculate the lengths of the sides.
Tips
- Forgetting to substitute values correctly when finding intersection points.
- Mixing up coordinates when applying the distance formula.
- Not checking if the intersection points yield a valid triangle (i.e., they must not be collinear).
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