and 
Then 
is (A) 12 (B) 
(C) 
(D) 
The correct option is OPTION(D) Given: Area of triangle = Modulus of = 35 Modulus of = 35 Taking positive sign, Taking negative sign,
and 
using determinants. (ii) Find equation of line joining 
and 
using determinants.Let $A(x, y)$ be any vertex of a triangle. All points are on one line if area of triangle is zero. $\frac{1}{2}[x(2-6)-y(1-3)+1(6-6)]=0$ $-4 x+2 y=0$ $y=2 x$ Which is equation of line. (ii) Let...
if area of triangle is 4 sq. units and vertices are (i) 
(ii) 
(i) $\frac{1}{2}[k(0-2)-0+1(8-0)]=4$ $1 / 2(-2 k+4)=4$ $-k+4=4$ Now: $-k+4=\pm 4$ $-k+4=4$ and $-k+4=-4$ $\mathrm{k}=0$ and $\mathrm{k}=8$ $\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} &...
are collinear.Points are collinear if area of triangle is equal to zero. i.e. Area of triangle $=0$ Area of Triangle $=\frac{1}{2}\left|\begin{array}{ccc}a & b+c & 1 \\ b & c+a & 1 \\ c & a+b...
Area $=\frac{1}{2}\left|\begin{array}{ccc}-2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1\end{array}\right|$ $=\frac{1}{2}[-2(10)+3(4)-22]$ $=15$ sq. Units
(ii) 
Formula for Area of triangle: $\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|$ (i)...