Question

Using determinant, find the area of the triangle whose vertices are

(− 3, 5), (3, − 6) and (7, 2).

Answer 1

hey here is ur answer

see the attachment above

I hope this helps you

Answer 2

We have to find the area of the triangle whose vertices are (-3 , 5), (3 , -6), (7 , 2).

(Method 1:

We know that the area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3), is given by the expression  [x1*(y2–y3) + x2*(y3–y1) + x3*(y1–y2)]/2 ..........................................................(1)

Then, area of the triangle will be [-3*( -6 -2) + 3*(2 - 5) + 7*(5 + 6)]/2

                                                       = 46 square-unit)

Method 2: (according to the question)

we can use determinant method to find the are of the triangle. We can write (1) as

2*area =

$\left[\begin{array}{ccc}-3&5&1\\3&-6&1\\7&2&1\end{array}\right]$

= $\left[\begin{array}{ccc}-3&5&1\\6&-11&0\\10&-3&0\end{array}\right]$

= -18 + 110

= 92 (Since the area is a positive quantity, we always take the absolute value of the determinant)

Then, area = 46 square-unit.(answer)