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Area of triangle formed by straight line...

Area of triangle formed by straight lines `x +y = 4, 2x - y =2 and x -2 = 0` is

A

4 sq. unit

B

9 sq. unit

C

`7/2` sq. unit

D

None of these

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To find the area of the triangle formed by the lines \(x + y = 4\), \(2x - y = 2\), and \(x - 2 = 0\), we will follow these steps: ### Step 1: Find the intersection points of the lines 1. **Line 1:** \(x + y = 4\) - If \(x = 0\), then \(y = 4\) (Point A: (0, 4)) - If \(y = 0\), then \(x = 4\) (Point B: (4, 0)) 2. **Line 2:** \(2x - y = 2\) - If \(x = 0\), then \(y = -2\) (Point C: (0, -2)) - If \(y = 0\), then \(2x = 2 \Rightarrow x = 1\) (Point D: (1, 0)) 3. **Line 3:** \(x - 2 = 0\) - This line is vertical at \(x = 2\). ### Step 2: Find the intersection of the first two lines To find the intersection of \(x + y = 4\) and \(2x - y = 2\), we can solve these equations simultaneously. 1. From \(x + y = 4\), we can express \(y\) in terms of \(x\): \[ y = 4 - x \] 2. Substitute \(y\) in the second equation: \[ 2x - (4 - x) = 2 \] \[ 2x - 4 + x = 2 \] \[ 3x - 4 = 2 \Rightarrow 3x = 6 \Rightarrow x = 2 \] 3. Substitute \(x = 2\) back into \(y = 4 - x\): \[ y = 4 - 2 = 2 \] So, the intersection point is (2, 2). ### Step 3: Identify the vertices of the triangle The vertices of the triangle formed by the lines are: - Point A: (0, 4) - Point B: (4, 0) - Point C: (2, 2) ### Step 4: Calculate the area of the triangle The area \(A\) of a triangle given vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula: \[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] Substituting the points: - \((x_1, y_1) = (0, 4)\) - \((x_2, y_2) = (4, 0)\) - \((x_3, y_3) = (2, 2)\) We get: \[ A = \frac{1}{2} \left| 0(0 - 2) + 4(2 - 4) + 2(4 - 0) \right| \] \[ A = \frac{1}{2} \left| 0 + 4(-2) + 2(4) \right| \] \[ A = \frac{1}{2} \left| 0 - 8 + 8 \right| = \frac{1}{2} \left| 0 \right| = 0 \] ### Conclusion Since the area calculated is 0, this indicates that the lines do not form a triangle. Therefore, the answer is **none of these**. ---
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LUCENT PUBLICATION-GRAPHICAL SOLUTION OF LINEAR EQUATION -EXERCISE-3A
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  4. Area of quadrilaeral formed by straight lines x + y = 2, 3x + 4y = 24 ...

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  5. A linear equation 3x + 4y = 24, intersects x-axis and y-axis respectiv...

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  11. Area of quadrilateral formed by striahgt lines x =1, x = 3, y =2 and ...

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  12. Area of triangle formed by straight lines x = 0, x + 2y = 0 and y = 1

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  13. Area of triangle formed by straight lines x - y = 0, x + y = 0 and 2 x...

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  14. Area enclosed by equation y = |x| - 5 with x-axis is

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  15. Area enclosed by the equation |x| + |y|= 4 is

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  16. Area enclosed by equation y = |x| -1 and y =1 - |x| is

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  17. Which of the following system of equations has unique solutions

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  19. Which of the following sytem of equations doesnot have a solution ?

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