Question

Find the area of triangle ABC bounded by:
5x −2y−10 =0, x−y−9=0, 3x−4y−6=0, usingintegrationmethod.

Answer 1

Question:
Find the area bounded by the lines:
1. \( 5x - 2y - 10 = 0 \)
2. \( x - y - 9 = 0 \)
3. \( 3x - 4y - 6 = 0 \)
using the method of integration.

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Solution:
Step 1: Convert the equations to slope-intercept form:
Line 1: \( 5x - 2y - 10 = 0 \Rightarrow y = \frac{5x - 10}{2} \)
Line 2: \( x - y - 9 = 0 \Rightarrow y = x - 9 \)
Line 3: \( 3x - 4y - 6 = 0 \Rightarrow y = \frac{3x - 6}{4} \)
Step 2: Find the points of intersection:
(i) Line 1 and Line 2:
\[ \frac{5x - 10}{2} = x - 9 \Rightarrow 5x - 10 = 2x - 18 \Rightarrow 3x = -8 \Rightarrow x = -\frac{8}{3} \] \[ y = x - 9 = -\frac{8}{3} - 9 = -\frac{35}{3} \] So, point \( A\left(-\frac{8}{3}, -\frac{35}{3}\right) \)
(ii) Line 2 and Line 3:
\[ x - 9 = \frac{3x - 6}{4} \Rightarrow 4x - 36 = 3x - 6 \Rightarrow x = 30,\ y = 30 - 9 = 21 \] So, point \( B(30, 21) \)
(iii) Line 1 and Line 3:
\[ \frac{5x - 10}{2} = \frac{3x - 6}{4} \Rightarrow 4(5x - 10) = 2(3x - 6) \Rightarrow 20x - 40 = 6x - 12 \Rightarrow 14x = 28 \Rightarrow x = 2 \] \[ y = \frac{5(2) - 10}{2} = 0 \] So, point \( C(2, 0) \)
Step 3: Set up integration to find area:
Between \( x = 2 \) and \( x = 30 \), the curves are:
- Upper curve: \( y = x - 9 \)
- Lower curve: \( y = \frac{3x - 6}{4} \)
Area = \[ \int_{2}^{30} \left[ (x - 9) - \frac{3x - 6}{4} \right] dx = \int_{2}^{30} \left[ \frac{4x - 36 - 3x + 6}{4} \right] dx = \int_{2}^{30} \left[ \frac{x - 30}{4} \right] dx \] \[ = \frac{1}{4} \int_{2}^{30} (x - 30)\ dx \]
Step 4: Integrate:
\[ \int_{2}^{30} (x - 30)\ dx = \left[ \frac{x^2}{2} - 30x \right]_{2}^{30} = \left( \frac{900}{2} - 900 \right) - \left( \frac{4}{2} - 60 \right) = (450 - 900) - (2 - 60) = -450 + 58 = -392 \] Final Answer: Take absolute value:
\[ \boxed{392 \text{ square units}} \]