Find the coordinates of the vertices of a triangle, the equations of whose sides are :

$(i) x + y - 4 = 0, 2 x - y + 3 = 0 a n d x - 3 y + 2 = 0$ $(i i) y (t_{1} + t_{2}) = 2 x + 2 a t_{1} t_{2}, y (t_{2} + t_{3}) = 2 x + 2 a t_{2} t_{3} a n d, y (t_{3} + t_{1}) = 2 x + 2 a t_{1} t_{3} .$

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Solution

$(i) The point of intersection of two sides will give the vertex$

$x + y - 4 = 0 . . . (1)$

$2 x - y + 3 = 0 . . . (2)$

$x - 3 y + 2 = 0 . . . (3)$

$Solving (1) and (2)$

$x + y = 4$

$y = 4 - x$

$Putting y in (2)$

$2 x - (4 - x) + 3 = 0$

$2 x - 4 + x + 3 = 0$

$3 x - 1 = 0$

$x = \frac{1}{3}$

$Putting x in (1)$

$\frac{1}{3} + y - 4 = 0$

$y = 4 - \frac{1}{3} = \frac{11}{3}$

$∴ One vertex is (\frac{1}{3}, \frac{11}{3})$

$Solving (2) and (3), we get$

$y = 2 x + 3 and putting in (3)$

$x - 3 y + 2 = 0$

$x - 3 (2 x + 3) + 2 = 0$

$x - 6 x - 9 + 2 = 0$

$- 5 x = + 7$

$x = \frac{- 7}{5}$

$\Rightarrow y = 2 x + 3 = 2 (\frac{- 7}{5}) + 3 = \frac{- 14}{5} + = \frac{1}{5}$

$∴ Second vertex is (\frac{- 7}{5}, \frac{1}{5})$

$For find vertex$

$x + y - 4 = 0$

$x - 3 y + 2 = 0$

$x = 4 - y$

$4 - y - 3 y + 2 = 0$

$4 - 4 y + 2 = 0$

$- 4 y = - 6$

$y = \frac{3}{2}$

$\Rightarrow x = 4 - y$

$4 - \frac{3}{2}$

$\frac{8 - 3}{2} = \frac{5}{2}$

$∴ Third vertex is (\frac{5}{2}, \frac{3}{2})$

$Hence co-ordinates of the vertices of the triangle are (\frac{1}{3}, \frac{11}{3}), (\frac{- 7}{5}, \frac{1}{5}) a n d (\frac{5}{2}, \frac{3}{2})$

$(i i) y (t_{1} + t_{2}) = 2 x + 2 a t_{1} t_{2}, y (t_{2} + t_{3}) = 2 x + 2 a t_{2} t_{3} a n d, y (t_{3} + t_{1}) = 2 x + 2 a t_{1} t_{3}$

$2 x - y (t_{1} + t_{2}) + 2 a t_{1} t_{2} = 0 . . . (1)$

$2 x - y (t_{2} + t_{3}) + 2 a t_{2} t_{3} = 0 . . . (2)$

$2 x - y (t_{3} + t_{1}) + 2 a t_{1} t_{3} = 0 . . . (3)$

$Solving (1) and (2) using cross-multiplication method:$

$\frac{x}{- 2 a t_{2} t_{3} (t_{1} + t_{2}) + 2 a t_{1} t_{2} (t_{2} + t_{3})}$

$\frac{y}{4 a t_{1} t_{2} - 4 a t_{2} t_{3}} = \frac{1}{- 2 (t_{2} + t_{3}) + 2 (t_{1} + t_{2})}$

$\Rightarrow \frac{x}{2 a t_{2}^{2} (t_{1} - t_{3})} = \frac{y}{4 a t_{2} (t_{1} - t_{3})} = \frac{1}{2 (t_{1} - t_{3})}$

$x = a t_{2}^{2}, y = 2 a t_{2}$

$Solving (1) and (3) using cross-multiplication method:$

$\frac{x}{- 2 a t_{1} t_{3} (t_{1} + t_{2}) + 2 a t_{1} t_{2} (t_{3} + t_{1})} = \frac{y}{4 a t_{1} t_{2} - 4 a t_{1} t_{3}} = \frac{1}{- 2 (t_{3} + t_{1}) + 2 (t_{1} + t_{2})}$

$\Rightarrow \frac{x}{2 a t_{1}^{2} (t_{2} - t_{3})} = \frac{y}{4 a t_{1} (t_{2} - t_{3})} = \frac{1}{2 (t_{2} - t_{3})}$

$\Rightarrow x = a t_{1}^{2}, y = 2 a t_{1}$

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