4.4 Q-14

Question Statement

Find the area of the region bounded by the triangle whose sides are represented by the following equations:

1. $7x - y - 10 = 0$
2. $10x + y - 41 = 0$
3. $3x + 2y + 3 = 0$

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Background and Explanation

In this problem, we are tasked with finding the area of a triangle formed by the intersections of three lines. To do this, we first need to determine the points of intersection of the given lines. Once we have the vertices of the triangle, we can apply the area formula for a triangle defined by three points in a coordinate plane.

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Solution

Step 1: Find the intersection of the lines

We will find the intersections of the lines pairwise. Let’s start with the first two lines:

1. Intersection of Line 1 and Line 2:

The equations are:

$$7x - y - 10 = 0 \quad \text{(1)}$$ $$10x + y - 41 = 0 \quad \text{(2)}$$

Add equations (1) and (2):

$$\frac{x}{(-1)(-41)-(1)(-10)} = \frac{-y}{(7)(-41)-(10)(-10)} = \frac{1}{(7)(1)-(10)(-1)}$$

Simplifying gives:

$$\frac{x}{41 + 40} = \frac{y}{187} = \frac{1}{7 + 10}$$

Thus, solving for $x$ and $y$:

$$x = \frac{51}{17} = 3, \quad y = \frac{187}{17} = 11$$

So the intersection point of Line 1 and Line 2 is $(3, 11)$.

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2. Intersection of Line 2 and Line 3:

The equations are:

$$10x + y - 41 = 0 \quad \text{(2)}$$ $$3x + 2y + 3 = 0 \quad \text{(3)}$$

Add equations (2) and (3):

$$\frac{x}{(1)(3)-(2)(-41)} = \frac{-y}{(10)(3)-(3)(-41)} = \frac{1}{(10)(2)-(3)(1)}$$

Simplifying gives:

$$\frac{x}{85} = \frac{y}{-153} = \frac{1}{17}$$

Thus, solving for $x$ and $y$:

$$x = \frac{85}{17} = 5, \quad y = \frac{153}{17} = -9$$

So the intersection point of Line 2 and Line 3 is $(5, -9)$.

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3. Intersection of Line 1 and Line 3:

The equations are:

$$7x - y - 10 = 0 \quad \text{(1)}$$ $$3x + 2y + 3 = 0 \quad \text{(3)}$$

Add equations (1) and (3):

$$\frac{x}{(-1)(3)-(2)(-10)} = \frac{-y}{(7)(3)-(3)(-10)} = \frac{1}{(7)(2)-(3)(-1)}$$

Simplifying gives:

$$\frac{x}{17} = \frac{y}{-51} = \frac{1}{17}$$

Thus, solving for $x$ and $y$:

$$x = \frac{17}{17} = 1, \quad y = \frac{-51}{17} = -3$$

So the intersection point of Line 1 and Line 3 is $(1, -3)$.

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Step 2: Calculate the Area of the Triangle

Now that we have the three vertices of the triangle $A(3, 11)$, $B(5, -9)$, and $C(1, -3)$, we can calculate the area using the formula for the area of a triangle with vertices at $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

$$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$

Substituting the coordinates $A(3, 11)$, $B(5, -9)$, and $C(1, -3)$:

$$\text{Area} = \frac{1}{2} \left| 3(-9 + 3) - 11(5 - 1) + 1(-15 + 9) \right|$$

Simplifying each term:

$$= \frac{1}{2} \left| 3(-6) - 11(4) + 1(-6) \right|$$ $$= \frac{1}{2} \left| -18 - 44 - 6 \right|$$ $$= \frac{1}{2} \times (-68)$$ $$= 34$$

Thus, the area of the triangle is $\boxed{34}$.

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Key Formulas or Methods Used

• Area of a Triangle Given Vertices:

$$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$

• Solving Systems of Linear Equations: Use substitution or elimination to find the intersection points of the lines.

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Summary of Steps

1. Solve for the intersection points of the lines pairwise.
2. Find the coordinates of the vertices of the triangle.
3. Use the area formula for a triangle with given vertices to calculate the area.