KCET · Maths · Straight Lines
The area of the triangle with vertices \((3, 8), (-4, 2)\) and \((5, 1)\) is \(\dfrac{P}{4}\), then the value of \(P\) is
- A \(\dfrac{61}{2}\)
- B \(\dfrac{2}{61}\)
- C \(122\)
- D \(\dfrac{1}{122}\)
Answer & Solution
Correct Answer
(C) \(122\)
Step-by-step Solution
Detailed explanation
The area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by \(\dfrac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\).
Substituting the given vertices \((3, 8)\), \((-4, 2)\), and \((5, 1)\):
Area \(= \dfrac{1}{2} |3(2 - 1) - 4(1 - 8) + 5(8 - 2)|\)
Area \(= \dfrac{1}{2} |3(1) - 4(-7) + 5(6)|\)
Area \(= \dfrac{1}{2} |3 + 28 + 30| = \dfrac{61}{2}\)
Given that the area is \(\dfrac{P}{4}\):
\(\dfrac{P}{4} = \dfrac{61}{2}\)
\(P = 122\)
Answer: \(122\)
Substituting the given vertices \((3, 8)\), \((-4, 2)\), and \((5, 1)\):
Area \(= \dfrac{1}{2} |3(2 - 1) - 4(1 - 8) + 5(8 - 2)|\)
Area \(= \dfrac{1}{2} |3(1) - 4(-7) + 5(6)|\)
Area \(= \dfrac{1}{2} |3 + 28 + 30| = \dfrac{61}{2}\)
Given that the area is \(\dfrac{P}{4}\):
\(\dfrac{P}{4} = \dfrac{61}{2}\)
\(P = 122\)
Answer: \(122\)
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