Chapter 1.1, Problem 58E

(a)

To determine

To find:The equation for the line $M$ that is perpendicular to given line $L$ and passes through point $P\left(a,b\right)$ .

Answer to Problem 58E

The equation for the line $M$ that is perpendicular to given line $L$ and passes through point $P\left(a,b\right)$ is $y=\frac{B}{A}\left(x-a\right)+b$ .

Explanation of Solution

Given information:The equation of line $L$ is $Ax+By=C$ and the point is $P\left(a,b\right)$ .

Calculation:

Rewrite the given equation of line $L$ in slope-intercept form as:

  $\begin{array}{c}By=-Ax+C\\ y=-\frac{A}{B}x+\frac{C}{B}\end{array}$

So, the slope of the line $L$ is $m_1=-\frac{A}{B}$ .

The condition for two perpendicular lines is $m_1{m}_2=-1$ . So, the slope of its perpendicular line can be calculated as:

  $\begin{array}{c}{m}_2=-\frac{1}{m_1}\\ =-\frac{1}{-\frac{A}{B}}\\ =\frac{B}{A}\end{array}$

The equation of its perpendicular line that passes through the point $P\left(a,b\right)$ is:

  $y=\frac{B}{A}\left(x-a\right)+b$

Therefore, the equation for the line $M$ that is perpendicular to given line $L$ and passes through point $P\left(a,b\right)$ is $y=\frac{B}{A}\left(x-a\right)+b$ .

(b)

To determine

To find:The coordinates of intersection point $Q$ of line $M$ and line $L$ .

Answer to Problem 58E

The coordinates of the point $Q$ are $\left(\frac{a{B}^2+CA-bAB}{B^2+A^2},\frac{b{A}^2+CB-aAB}{B^2+A^2}\right)$ .

Explanation of Solution

Given information:The equation of line $L$ is $Ax+By=C$ and the point is $P\left(a,b\right)$ .

Calculation:

From part (a), the equation for the line $M$ is $y=\frac{B}{A}\left(x-a\right)+b$ .

The given equation of line $L$ is $y=-\frac{A}{B}x+\frac{C}{B}$ .

Substitute $\frac{B}{A}\left(x-a\right)+b$ for $y$ in the equation of line $L$ and solve.

  $\begin{array}{c}\frac{B}{A}\left(x-a\right)+b=-\frac{A}{B}x+\frac{C}{B}\\ \frac{B}{A}x-a\frac{B}{A}+b=-\frac{A}{B}x+\frac{C}{B}\\ \frac{B}{A}x+\frac{A}{B}x=\frac{C}{B}+a\frac{B}{A}-b\\ \frac{\left(B^2+A^2\right)}{AB}x=\frac{CA+a{B}^2-bAB}{AB}\end{array}$

Simplify further.

  $\begin{array}{c}\left(B^2+A^2\right)x=CA+a{B}^2-bAB\\ x=\frac{CA+a{B}^2-bAB}{B^2+A^2}\end{array}$

Now, substitute the value of $x$ in the equation $y=-\frac{A}{B}x+\frac{C}{B}$ .

  $\begin{array}{c}y=-\frac{A}{B}\left(\frac{CA+a{B}^2-bAB}{B^2+A^2}\right)+\frac{C}{B}\\ =\frac{-C{A}^2-aA{B}^2+b{A}^{2}B+C{B}^2+C{A}^2}{B\left(B^2+A^2\right)}\\ =\frac{B\left(b{A}^2+CB-aAB\right)}{B\left(B^2+A^2\right)}\\ =\frac{b{A}^2+CB-aAB}{A^2+B^2}\end{array}$

Therefore, the coordinates of the point $Q$ are $\left(\frac{a{B}^2+CA-bAB}{B^2+A^2},\frac{b{A}^2+CB-aAB}{B^2+A^2}\right)$ .

(c)

To determine

To find:The distance from point $P$ to point $Q$ .

Answer to Problem 58E

The distance from point $P$ to point $Q$ is $\frac{\left|C-bB-aA\right|}{\sqrt{A^2+B^2}}$ .

Explanation of Solution

Given information:The point is $P\left(a,b\right)$ .

Calculation:

From part (b), the coordinates of the point $Q$ are $\left(\frac{a{B}^2+CA-bAB}{B^2+A^2},\frac{b{A}^2+CB-aAB}{B^2+A^2}\right)$ .

The formula to find the distance between two points is given by,

  $\text{distance}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The distance from point $P$ to point $Q$ can be calculated as:

  $\begin{array}{c}\text{distance}=\sqrt{{\left(\frac{a{B}^2+CA-bAB}{B^2+A^2}-a\right)}^2+{\left(\frac{b{A}^2+CB-aAB}{B^2+A^2}-b\right)}^2}\\ =\sqrt{{\left(\frac{a{B}^2+CA-bAB-a{B}^2-a{A}^2}{B^2+A^2}\right)}^2+{\left(\frac{b{A}^2+CB-aAB-b{B}^2-b{A}^2}{B^2+A^2}\right)}^2}\\ =\sqrt{{\left(\frac{CA-bAB-a{A}^2}{B^2+A^2}\right)}^2+{\left(\frac{CB-aAB-b{B}^2}{B^2+A^2}\right)}^2}\\ =\sqrt{\frac{\left(A^2+B^2\right){\left(C-bB-aA\right)}^2}{{\left(A^2+B^2\right)}^2}}\end{array}$

Simplify further.

  $\begin{array}{c}\text{distance}=\sqrt{\frac{{\left(C-bB-aA\right)}^2}{A^2+B^2}}\\ =\frac{\left|C-bB-aA\right|}{\sqrt{A^2+B^2}}\end{array}$

Therefore, the distance from point $P$ to point $Q$ is $\frac{\left|C-bB-aA\right|}{\sqrt{A^2+B^2}}$ .