Find area of triangle with vertices at point given in each of following: (1,0),(6,0),(4,3)

English

Gujarati

Hindi

3 and 4 .Determinants and Matrices

easy

Find area of the triangle with vertices at the point given in each of the following: $(1,0),(6,0),(4,3)$

A

$\frac{11}{2}$ square units

B

$\frac{17}{2}$ square units

C

$\frac{15}{2}$ square units

D

$\frac{19}{2}$ square units

Solution

The area of the triangle with vertices $(1,0),(6,0),(4,3)$ is given by the relation,

$\Delta=\frac{1}{2}\left|\begin{array}{lll}1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1\end{array}\right|$

$=\frac{1}{2}[1(0-3)-0(6-4)+1(18-0)]$

$=\frac{1}{2}[-3+18]=\frac{15}{2}$ square units

Standard 12

Mathematics

Share

Similar Questions

The determinant $\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&3&6\end{array}\,} \right|$ is not equal to

medium

The values of $x $ in the following determinant equation, $\left| {\,\begin{array}{*{20}{c}}{a + x}&{a – x}&{a – x}\\{a – x}&{a + x}&{a – x}\\{a – x}&{a – x}&{a + x}\end{array}\,} \right| = 0$ are

easy

If the system of linear equations $2 \mathrm{x}+2 \mathrm{ay}+\mathrm{az}=0$ ; $2 x+3 b y+b z=0$ ; $2 \mathrm{x}+4 \mathrm{cy}+\mathrm{cz}=0$ ; where $a, b, c \in R$ are non-zero and distinct; has a non-zero solution, then

hard

(JEE MAIN-2020)

$\left| {\,\begin{array}{*{20}{c}}{19}&{17}&{15}\\9&8&7\\1&1&1\end{array}\,} \right| = $

easy

Evaluate the determinants

$\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|$

easy