From the coordinates of the triangle formed by the line 4S minus 5 y - 20 is equal to zero and 3 x + 5 y - 15 is equal to zero and y axis find the area of triangle
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Taking equation (1)
4x−5y−20=0
x 0 5
Y −4 0
tAKING EQUATION (2)
3X+5Y−15=0
X 0 5
Y 3 0
⇒ Solution is given by the point where 2 times intersect each others i.e (5,0)
∴ x=5 y=0 (required solution)
⇒ Vertices of triangle ABC=(0,3)(5,0)(0,−4)
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