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Question
Find the coordinates of the vertices of a triangle, the equations of whose sides are x + y − 4 = 0, 2x − y + 3 = 0 and x − 3y + 2 = 0.
Solution
x + y − 4 = 0, 2x − y + 3 = 0 and x − 3y + 2 = 0
x + y − 4 = 0 ... (1)
2x − y + 3 = 0 ... (2)
x − 3y + 2 = 0 ... (3)
Solving (1) and (2) using cross-multiplication method:
\[\frac{x}{3 - 4} = \frac{y}{- 8 - 3} = \frac{1}{- 1 - 2}\]
\[ \Rightarrow x = \frac{1}{3}, y = \frac{11}{3}\]
Solving (1) and (3) using cross-multiplication method:
\[\frac{x}{2 - 12} = \frac{y}{- 4 - 2} = \frac{1}{- 3 - 1}\]
\[ \Rightarrow x = \frac{5}{2}, y = \frac{3}{2}\]
Similarly, solving (2) and (3) using cross-multiplication method:
\[\frac{x}{- 2 + 9} = \frac{y}{3 - 4} = \frac{1}{- 6 + 1}\]
\[ \Rightarrow x = - \frac{7}{5}, y = \frac{1}{5}\]
Hence, the coordinates of the vertices of the triangle are \[\left( \frac{1}{3}, \frac{11}{3} \right)\], \[\left( \frac{5}{2}, \frac{3}{2} \right)\] and \[\left( - \frac{7}{5}, \frac{1}{5} \right)\].
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