Area of triangles formed by the lines $x + 2 y - 3 = 0, x + 4 y - 5 = 0$ $x + 5 y - 1 = 0$

49 / 2

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80 / 3

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27 / 3

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7 / 2

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Solution

The correct option is B $80 / 3$
Given lines are
$l_{1} : x + 2 y - 3 = 0$
$l_{2} : x + 4 y - 5 = 0$
$l_{3} : x + 5 y - 1 = 0$

Intersection of line 1 and 2 gives A(1,1)
Intersection of line 2 and 3 gives B(21,-4)
Intersection of line 3 and 1 gives C( $\frac{13}{3}, \frac{- 2}{3}$ )

Area of triangle formed by A(1,1) ,B(21,-4) and C( $\frac{13}{3}, \frac{- 2}{3})$

= $\frac{1}{2} | 1 (- 4 + \frac{2}{3}) - 21 (\frac{- 2}{3} - 1) + \frac{13}{3} (1 - (- 4)) |$

= $= \frac{1}{2} | \frac{- 10}{3} + 35 + \frac{65}{3} | \frac{1}{2} \times \frac{160}{3} = \frac{80}{3} s q . u n i t s$

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Area of triangles formed by the lines x+2y−3=0,x+4y−5=0x+5y−1=0

Area of triangles formed by the lines $x + 2 y - 3 = 0, x + 4 y - 5 = 0$ $x + 5 y - 1 = 0$