Question

Find the area of the triangle formed by the sides. $x=0,x+2y=5,3x–y=1$

• A. $\frac{7}{4}$
• B. $\frac{1}{4}$
• C. $\frac{3}{7}$
• D. $\frac{5}{7}$

Answer 1

The correct option is A

$\frac{7}{4}$

Solution :Given sides are,

$x=0$

$x+2y–5=0$

$3x–y–1=0$

First of all lets recollect the formula for finding the area of the triangle when its sides are given.

$△=\frac{1}{2C_1{C}_2{C}_3}\cdot {\left|\begin{array}{ccc}{a}_1& b_1& C_1\\ a_2& b_2& C_2\\ a_3& b_3& C_3\end{array}\right|}^2$

Where $C_1,C_2,C_3$ are cofactors of the elements $c_1,c_2,c_3$ respectively. Before doing that lets consider the coefficient matrix.

$\left[\begin{array}{ccc}{a}_1& b_1& C_1\\ a_2& b_2& C_2\\ a_3& b_3& C_3\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 1& 2& -5\\ 3& -1& -1\end{array}\right]$

$C_1=a_2{b}_3–b_2{a}_3=–1–6=–7$

$C_2=b_1{a}_3–a_1{b}_3=0–(-1)=1$

$C_3=a_1{b}_2–b_1{a}_2=2$

Also

$|A|=1(-2-5)=-7$

$\therefore Area=\frac{1}{2(-7)1\cdot 2}\cdot (-7{)}^2$

$=|\frac{-7}{4}|=\frac{7}{4}\text{sq. units.}$