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| 1. |
in each of the following find value of k 1- (7,-2),(5,1),(3,k) |
| Answer» (7, –2), (5, 1), (3, k)Area of the triangle{tex}= \\frac{1}{2}{/tex}[7(1 – k) + 5(k – (–2)) + 3(–2 – 1)]{tex}= \\frac{1}{2}{/tex}[7 – 7k + 5k + 10\xa0– 9]{tex}= \\frac{1}{2}{/tex}[8 – 2k] = 4– kIf the points are collinear, then area of the triangle = 0{tex}\\Rightarrow{/tex} 4 – k = 0{tex}\\Rightarrow{/tex} k = 4 | |