Question

the area of triangle formed by the lines 2x+3y=12 , x-y-1=0 and x=0 is

Answer 1

Answer:

$\red {Area \:of \:the \: \triangle }\green {= 5\:square \:units }$

Step-by-step explanation:

From above graph ,

i) Red line shows x = 0,

ii ) Blue line shows 2x+3y = 12

iii) Green line shows x-y-1 = 0

Intersecting points of these line are (1,0), (0,4) and (3,2)

$Let \:A(1,0) = (x_{1},y_{1}),\\\:B(0,4) = (x_{2},y_{2}),\\\:C(3,2) = (x_{3},y_{3}),$

$\boxed {\pink { Area \:of \:a \: triangle = \frac{1}{2}|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|}}$

$= \frac{1}{2}|1(4-2)+0(2-0)+3(0-4)|$

$=\frac{1}{2}|2+0+3(-4)|$

$=\frac{1}{2}|2-12|$

$=\frac{1}{2}|-10|$

$=\frac{1}{2}\times 10\\=5\: square\:units$

Therefore.,

$\red {Area \:of \:the \: \triangle }\green {= 5\:square \:units }$

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Answer 2

Answer:

Step-by-step explanation: