Draw the graphs of the following equations:
$2 x - 3 y + 6 = 0$
$2 x + 3 y - 18 = 0$
$y - 2 = 0$
Find the vertices of the triangle so obtained. Also, find the area of the triangle.

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Solution

Given,

$2 x - 3 y + 6 = 0$

$2 x - 3 y = - 6$

Divide the equation by $- 6$

$\Rightarrow - \frac{x}{3} + \frac{y}{2} = 1$

$2 x + 3 y - 18 = 0$

$2 x + 3 y = 18$

Divide the equation by 18

$\Rightarrow \frac{x}{9} + \frac{y}{6} = 1$

$y - 2 = 0$

$\Rightarrow y = 2$

Vertices of trianlge formed by the intersection of these lines is given by:

$a (3, 4), b (0, 2), c (6, 2)$

area of trianlge

$A = \frac{a_{x} (b_{y} - c_{y}) + b_{x} (c_{y} - a_{y}) + c_{x} (a_{y} - b_{y})}{2}$

$= \frac{3 (2 - 2) + 0 (2 - 4) + 6 (4 - 2)}{2}$

$= \frac{6 \times 2}{2}$

$= 6$

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Draw the graphs of the following equations:2x−3y+6=02x+3y−18=0y−2=0Find the vertices of the triangle so obtained. Also, find the area of the triangle.

Draw the graphs of the following equations:
$2 x - 3 y + 6 = 0$
$2 x + 3 y - 18 = 0$
$y - 2 = 0$
Find the vertices of the triangle so obtained. Also, find the area of the triangle.