Question

Find the area of ΔABC, whose vertices are

A(1,-4), B (3,2) and C(-1,2)​

Answer 1

I don't understand your question

Answer 2

Answer:

Let BC be the base and AD be height

AB= ✓(x2-x1)²+(y2-y1)²

=>√(3-1)²+(2-(-4))²

=>√(2)²+(2+4)²

=>√4+6²

=>√4+36

=>√40

=>2√10

BC= √(x2-x1)²+(y2-y1)²

=>√(-1-3)²+(2-2)²

=>√(-4)²+0

=>√16+0

=>4

CA=√(x2-x1)²+(y2-y1)²

=>√(1-(-1))²+(2-(-4))²

=>√(1+1)²+(2+4)²

=>√(2)²+(6)²

=>√4+36

=>√40

=>2√10

Here sides AB and CA are equal therefore it's an isoceles triangle....

Area of isoceles triangle

1/2xbasexheight but we don't know the height..

We know that the height of an isoceles triangle divides it into two congruent triangles...

So two triangles ADB and ADC are formed

Now, Let's consider triangle ADB

Let AB=a

BD=b

and AD=c

hence a=2√10 and b=2(Because BC is 4 and is divided equally into two triangles by AD)

a²+b²=c²

2√10²+2²=c²

20+4=c²

24=c²

c=2√6

therefore AD=2✓6

now area

1/2xbasexheight

=>1/2x4x2√6

=>4√6

I am right but your question is wrong I guess...