Chapter 7.2, Problem 70E

Solution

a.

To prove: The equation is linear in x and y .

The statement has proven.

Given Information:

The matrix is defined as,

  $\left|\begin{array}{ccc}1& x& y\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0$

Calculation:

Consider the given information,

  $\left|\begin{array}{ccc}1& x& y\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0$

Now, find the given determinant.

  $\begin{array}{c}\left|\begin{array}{ccc}1& x& y\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0\\ 1\left[x_1{y}_2-y_1{x}_2\right]-x\left[y_2-y_1\right]+y\left[x_2-x_1\right]=0\\ y\left[x_2-x_1\right]=x\left[y_2-y_1\right]-\left[x_1{y}_2-y_1{x}_2\right]\\ y=x\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}-\frac{\left(x_1{y}_2-y_1{x}_2\right)}{\left(x_2-x_1\right)}\end{array}$

The above equation is linear equation.

Therefore, the given statement has proved.

b.

To prove: The coordinates $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ lie on the line in part (a).

The statement has proven.

Given Information:

The determinant is defined as,

  $\left|\begin{array}{ccc}1& x& y\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0$

Explanation:

Consider the given equations,

  $\left|\begin{array}{ccc}1& x& y\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0$

Substitute the first point in the equation.

  $\begin{array}{c}{y}_1=x_1\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}-\frac{\left(x_1{y}_2-y_1{x}_2\right)}{\left(x_2-x_1\right)}\left[\left(x,y\right)=\left(x_1,y_1\right)\right]\\ y_1=\frac{-x_1{y}_1+y_1{x}_2}{\left(x_2-x_1\right)}\\ y_1{x}_2-y_1{x}_1=-x_1{y}_1+y_1{x}_2\\ y_1{x}_1=x_1{y}_1\end{array}$

As the left side is equal to the right side.

Thus, the first point line on the equation. Now, check the other point.

  $\begin{array}{c}{y}_2=x_2\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}-\frac{\left(x_1{y}_2-y_1{x}_2\right)}{\left(x_2-x_1\right)}\left[\left(x,y\right)=\left(x_2,y_2\right)\right]\\ y_2=\frac{y_2{x}_2-x_1{y}_2}{\left(x_2-x_1\right)}\\ y_2{x}_2-y_2{x}_1=y_2{x}_2-x_1{y}_2\\ y_2{x}_2=y_2{x}_2\end{array}$

As the left side is equal to the right side.

Thus, the first point line on the equation. Now, check the other point.

Hence, the given statement has proved.

c.

To explain: The point $\left(x_3,y_3\right)$ lies on the line in part (a) by using a determinant.

The point will lie on the line, if the determinant become 0 for the given point.

Given Information:

The determinant is defined as,

  $\left|\begin{array}{ccc}1& x& y\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0$

Explanation:

Consider the given information,

  $\left|\begin{array}{ccc}1& x& y\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0$

Substitute the given point $\left(x_3,y_3\right)$ in the obtained determinant in part (a).

  $\begin{array}{c}\left|\begin{array}{ccc}1& x_3& y_3\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0\\ 1\left[x_1{y}_2-y_1{x}_2\right]-x_3\left[y_2-y_1\right]+y_3\left[x_2-x_1\right]=0\\ y_3\left[x_2-x_1\right]=x_3\left[y_2-y_1\right]-\left[x_1{y}_2-y_1{x}_2\right]\\ y_3=x_3\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}-\frac{\left(x_1{y}_2-y_1{x}_2\right)}{\left(x_2-x_1\right)}\end{array}$

If the right side come to equal to the right side then the point $\left(x_3,y_3\right)$ will lie on the line.

Therefore, the required result is explained above.

d.

To explain: The point $\left(x_3,y_3\right)$ lies on the line in part (a) by using a determinant.

The point will not lie on the line, if the determinant do not become 0 for the given point.

Given Information:

The determinant is defined as,

  $\left|\begin{array}{ccc}1& x& y\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0$

Explanation:

Consider the given information,

  $\left|\begin{array}{ccc}1& x& y\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0$

Substitute the given point $\left(x_3,y_3\right)$ in the obtained determinant in part (a).

  $\begin{array}{c}\left|\begin{array}{ccc}1& x_3& y_3\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0\\ 1\left[x_1{y}_2-y_1{x}_2\right]-x_3\left[y_2-y_1\right]+y_3\left[x_2-x_1\right]=0\\ y_3\left[x_2-x_1\right]=x_3\left[y_2-y_1\right]-\left[x_1{y}_2-y_1{x}_2\right]\\ y_3=x_3\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}-\frac{\left(x_1{y}_2-y_1{x}_2\right)}{\left(x_2-x_1\right)}\end{array}$

Here, if the condition $y_3\ne x_3\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}-\frac{\left(x_1{y}_2-y_1{x}_2\right)}{\left(x_2-x_1\right)}$ is true then the point $\left(x_3,y_3\right)$ will not lie one the line. The value of determinant cannot be zero for the point $\left(x_3,y_3\right)$ .

Therefore, the required result is explained above.