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)
III. (−2,−3),(3,2),(−1,−8)
(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,
\(\triangle\)=\(\frac{1}{2}\)\(\begin{vmatrix}1&0&1\\6&0&1\\4&3&1\end{vmatrix}\)
=\(\frac{1}{2}\)[1(0-3)-0(6-4)+1(18-0)]
=\(\frac{1}{2}\)[-3+18]=\(\frac{15}{2}\) square units
(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,
△=\(\frac{1}{2}\)\(\begin{vmatrix}2&7&1\\1&1&1\\10&8&1\end{vmatrix}\)
=\(\frac{1}{2}\) [2(1-8)-7(1-10)+1(8-10)]
=\(\frac{1}{2}\)[-14+63-2]
=\(\frac{1}{2}\)[-16+63]
=\(\frac{47}{2}\) square units
(iii) The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8)
is given by the relation,
△=\(\frac{1}{2}\)\(\begin{vmatrix}-2&-3&1\\3&2&1\\-1&-8&1\end{vmatrix}\)
=\(\frac{1}{2}\)[-2(2+8)+3(3+1)+1(-24+2)]
=\(\frac{1}{2}\)[-20+12-22]
=-\(\frac{30}{2}\)=-15
Hence, the area of the triangle is I-15I=15 square units.
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Evaluate the determinants in Exercises 1 and 2.
\(\begin{vmatrix}2&4\\-5&-1\end{vmatrix}\)
Evaluate the determinants in Exercises 1 and 2.
(i) \(\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}\)
(ii) \(\begin{vmatrix}x^2&-x+1&x-1\\& x+1&x+1\end{vmatrix}\)
Using properties of determinants,prove that:
\(\begin{vmatrix} x & x^2 & 1+px^3\\ y & y^2 & 1+py^3\\z&z^2&1+pz^3 \end{vmatrix}\)\(=(1+pxyz)(x-y)(y-z)(z-x)\)
Using properties of determinants,prove that:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)