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The area of the triangle with vertices a...

The area of the triangle with vertices at (-4,1),(1,2),(4,-3) is

A

14

B

16

C

15

D

None of these

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The correct Answer is:
To find the area of the triangle with vertices at points A(-4, 1), B(1, 2), and C(4, -3), we can use the formula for the area of a triangle given by the coordinates of its vertices: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] Where \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) are the coordinates of the vertices. ### Step 1: Assign the coordinates Let: - \(A(-4, 1) \Rightarrow (x_1, y_1) = (-4, 1)\) - \(B(1, 2) \Rightarrow (x_2, y_2) = (1, 2)\) - \(C(4, -3) \Rightarrow (x_3, y_3) = (4, -3)\) ### Step 2: Substitute the coordinates into the area formula Substituting the coordinates into the area formula: \[ \text{Area} = \frac{1}{2} \left| -4(2 - (-3)) + 1((-3) - 1) + 4(1 - 2) \right| \] ### Step 3: Simplify the expression Calculating each term step-by-step: 1. Calculate \(2 - (-3) = 2 + 3 = 5\) 2. Calculate \((-3) - 1 = -4\) 3. Calculate \(1 - 2 = -1\) Now substituting these values back into the area formula: \[ \text{Area} = \frac{1}{2} \left| -4(5) + 1(-4) + 4(-1) \right| \] ### Step 4: Calculate the products Calculating the products: 1. \(-4 \times 5 = -20\) 2. \(1 \times -4 = -4\) 3. \(4 \times -1 = -4\) Now substituting these results: \[ \text{Area} = \frac{1}{2} \left| -20 - 4 - 4 \right| \] ### Step 5: Combine the terms Combine the terms inside the absolute value: \[ -20 - 4 - 4 = -28 \] So, \[ \text{Area} = \frac{1}{2} \left| -28 \right| = \frac{1}{2} \times 28 = 14 \] ### Conclusion Thus, the area of the triangle is \(14\) square units.
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ML KHANNA-RECTANGULAR CARTESIAN CO-ORDINATE SYSTEM AND THE STRAIGHT LINE-SELF ASSESSMENT TEST
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  9. P and Q are points on the line joining A(-2,5) and B(3,1) such that AP...

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  11. The distance between the lines 3x+4y=9 and 6x+8y=15 is

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