Question

Find the area of triangle whode vertices are (4,3) (1,4) and (2,3)

Answer 1

Step-by-step explanation:

see here i will not give you the answer right now as I currently don't have a pen and but I am going to tell you the way to do it.

See starting with the Ques first check whether it is a right angled triangle or not. this you can do by slope form i.e., find the slope between the three given points one by one ( slope=(y2-y1)/(x2-x1)). if any two slope comes in the form as: M1*M2=-1. i.e. product of any two slope =-1 then it is a right angled triangle

Then if it is a right angled triangle then by distance formula find the length of each side then use area=1/2*base* height.

Then if it is not a right angled triangle then use herons formula.

Herons formula={√s(s-a)(s-b)(s-c)} remember all the things are in square root. Now s=(a+b+c)/2 where,

a,b,c are the length of the sides of the triangle.

Now if you know determinants then it is going to be a quite easy task .

I will tell u that way also (do only if know just how to find the value of the determinants).

area=1/2* take 3*3 determinant then take first row of determinants as 1 and then fill the other two rows with the coordinate given exactly in the same order as it is. and just solve the determinants which is a one line Ques. and you will get the ans.