The correct option is OPTION(D) Given: Area of triangle = Modulus of = 35 Modulus of = 35 Taking positive sign, Taking negative sign,
(i) Find equation of line joining $(1,2)$ and $(3,6)$ using determinants. (ii) Find equation of line joining $(3,1)$ and $(9,3)$ 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...
Find values of $k$ if area of triangle is 4 sq. units and vertices are (i) $(k, 0),(4,0),(0,2)$ (ii) $(-2,0),(0,4),(0, k)$
(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} &...
Show that points: $A(a, b+c), B(b, c+a), C(c, a+b)$ 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...
Find area of the triangle with vertices at the point given in each of the following:(iii) $(-2,-3),(3,2),(-1,-8)$
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
Find area of the triangle with vertices at the point given in each of the following: (i) $(1,0),(6,0),(4,3)$ (ii) $(2,7),(1,1),(10,8)$
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)...