InterviewSolution
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InterviewSolution
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The area of a triangle is 5. Two of its vertices are (2, 1) and (3, –2). The third vertex is (x, y) where y = x + 3. Find the co-ordinates of the third vertex. |
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Answer» Let A ≡ (x, y), B ≡ (2, 1), C ≡ (3, –2) Area of ΔABC = \(\frac{1}{2}\) |{x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2)}| = \(\frac{1}{2}\) | \(x\)(1 + 2) + 2(–2 – y) + 3(y – 1) | = \(\frac{1}{2}\) | 3\(x\) – 4 – 2y + 3y – 3 | = \(\frac{1}{2}\) | 3\(x\) + y – 7 | Given \(\frac{1}{2}\) | 3\(x\) + y – 7 | = 5 ⇒ | 3\(x\) + y – 7 | = 10 ⇒ 3\(x\) + y – 7 = 10 or –(3\(x\) + y – 7) = 10 ⇒ 3\(x\) + y = 17 or 3\(x\) + y = –3 Case I. 3\(x\) + y = 17. Also given y = \(x\) + 3 ∴ 3x + \(x\) + 3 = 17 ⇒ 4\(x\) = 14 ⇒ \(x\) = \(\frac{7}{2}\) ⇒ y = \(\frac{7}{2}\) + 3 = \(\frac{13}{2}\) Case II. 3\(x\) + y = –3, y = \(x\) + 3 ∴ 3\(x\) + \(x\) + 3 = –3 ⇒ 4\(x\) = – 6 ⇒ \(x\) = \(\frac{-3}{2}\) ⇒ y = \(\frac{-3}{2}\) + 3 = \(\frac{3}{2}\) ∴ Co-ordinates of A are \(\bigg(\frac{7}{2},\frac{13}{2}\bigg)\) or \(\bigg(\frac{-3}{2},\frac{3}{2}\bigg)\) |
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