Question

13. Find the area of a triangle with vertices (0,4), (0, 2) and (3,0)​

Answer 1

$\huge\underline\mathfrak{solution}$

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Answer 2

Step-by-step explanation:

let A(X1 , Y1)=(0,4)

B(X2,Y2)=(0,2)

C(X3,Y3)=(3,0)

area of triangle=1/2| X1(Y2-Y3)+X2(Y3-Y1)+ X3(Y1-Y2)|sq.units

∆=1/2|0(2-0)+0(0-4)+3(4-2)|

∆=1/2|0+0+6|

∆=1/2(6)

∆= 3 sq.units