Chapter 7, Problem 116RE

To determine

To calculate: The equation of the line passing through $\left(-0.8,0.2\right)\text{ and }\left(0.7,3.2\right)$.

The equation of line is $x+0.5y+0.9=0$.

Calculation:

Consider the points,

$\left(x_1,y_1\right)=\left(-0.8,0.2\right),\left(x_2,y_2\right)=\left(0.7,3.2\right)$

Substitute all the values in the standard determinant $\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$. This gives,

$\left|\begin{array}{ccc}x& y& 1\\ -0.8& 0.2& 1\\ 0.7& 3.2& 1\end{array}\right|$

The determinant of the matrix along row $1$ is,

$\left|\begin{array}{ccc}x& y& 1\\ -0.8& 0.2& 1\\ 0.7& 3.2& 1\end{array}\right|=a_{11}{C}_{11}+a_{12}{C}_{12}+a_{13}{C}_{13}$

From the matrix $a_{11}=x,a_{12}=y\text{ and }{a}_{13}=1$.

Substitute the value in the determinant,

$\left|\begin{array}{ccc}x& y& 1\\ -0.8& 0.2& 1\\ 0.7& 3.2& 1\end{array}\right|=x.C_{11}+y.C_{12}+1.C_{13}$

The coefficient of the cofactors $C_{11},C_{12}\text{ and }{C}_{13}$ are given as,

Compute the cofactor, $C_{11}$ as,

$C_{11}={\left(-1\right)}^{1+1}{M}_{11}$

To compute $M_{11}$, delete the first row and the first column,

$\begin{array}{c}{C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}0.2& 1\\ 3.2& 1\end{array}\right|\\ ={\left(-1\right)}^2\left|\begin{array}{cc}0.2& 1\\ 3.2& 1\end{array}\right|\\ =\left|\begin{array}{cc}0.2& 1\\ 3.2& 1\end{array}\right|\end{array}$

Now, expand using cofactors,

$\begin{array}{c}\left|\begin{array}{cc}0.2& 1\\ 3.2& 1\end{array}\right|=\left(\left(0.2\times 1\right)-\left(1\times 3.2\right)\right)\\ =0.2-3.2\\ =-3\end{array}$

To compute $M_{11}$, delete the first row and the first column,

$\begin{array}{c}{C}_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}-0.8& 1\\ 0.7& 1\end{array}\right|\\ ={\left(-1\right)}^3\left|\begin{array}{cc}-0.8& 1\\ 0.7& 1\end{array}\right|\\ =\left(-1\right)\left|\begin{array}{cc}-0.8& 1\\ 0.7& 1\end{array}\right|\end{array}$

Now, expand using cofactors,

$\begin{array}{c}\left|\begin{array}{cc}-0.8& 1\\ 0.7& 1\end{array}\right|=\left(\left(-0.8\times 1\right)-\left(1\times 0.7\right)\right)\\ =-0.8-0.7\\ =-1.5\end{array}$

To compute $M_{11}$, delete the first row and the first column,

$\begin{array}{c}{C}_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}-0.8& 0.2\\ 0.7& 3.2\end{array}\right|\\ ={\left(-1\right)}^4\left|\begin{array}{cc}-0.8& 0.2\\ 0.7& 3.2\end{array}\right|\\ =\left|\begin{array}{cc}-0.8& 0.2\\ 0.7& 3.2\end{array}\right|\end{array}$

Now, expand using cofactors,

$\begin{array}{c}\left|\begin{array}{cc}-0.8& 0.2\\ 0.7& 3.2\end{array}\right|=\left(\left(-0.8\times 3.2\right)-\left(0.2\times 0.7\right)\right)\\ =-2.56-0.14\\ =-2.7\end{array}$

Now, substitute the values in the determinant,

$\begin{array}{c}\left[\begin{array}{ccc}x& y& 1\\ -0.8& 0.2& 1\\ 0.7& 3.2& 1\end{array}\right]=x\cdot \left(-3\right)+y\cdot \left(-1.5\right)+1\cdot \left(-2.7\right)\\ 0=-3x-1.5y-2.7\\ 0=x+0.5y+0.9\end{array}$

The value of the determinant is $0$ since the points are collinear.

The equation of line is $x+0.5y+0.9=0$.