Chapter 0.1, Problem 58E

a.

To determine

To find: The equation of line $M$ through $P$ and perpendicular to $L$ .

Answer to Problem 58E

  $Bx-Ay=Ba-Ab$

Explanation of Solution

Given Information:

  $\begin{array}{l}P\left(a,b\right)\\ L:Ax+By=C\end{array}$

Concept used:

The equation of perpendicular line to the given equation $ax+by=c$ is,

  $bx-ay=k$

Calculation:

The required equation of line is perpendicular to $sAx+By=C$ . So, the equation of line is,

  $Bx-Ay=k$

The passing point is $P\left(a,b\right)$ .Put $x=a\text{ and }y=b$ into the equation,

  $Ba-Ab=k$

The required equation of line is, $Bx-Ay=Ba-Ab$

b.

To determine

To find: The point of interception of line $M$ and $L$ .

Answer to Problem 58E

  $Q\left(\frac{AC+B^{2}a-ABb}{A^2+B^2},\frac{BC+A^{2}b-ABa}{A^2+B^2}\right)$

Explanation of Solution

Given Information:

  $\begin{array}{l}P\left(a,b\right)\\ L:Ax+By=C\\ M:Bx-Ay=Ba-Ab\end{array}$

Calculation:

  $\begin{array}{c}Ax+By=C\\ Bx-Ay=Ba-Ab\end{array}$

Using elimination method, to eliminate the value of y. So, multiply first equation with $A$ and second equation with $B$ .

  $\begin{array}{l}\begin{array}{ccc}{A}^{2}x+ABy& =& AC\\ B^{2}x-ABy& =& B^{2}a-ABb\\ x\left(A^2+B^2\right)& =& AC+B^{2}a-ABb\end{array}\\ x=\frac{AC+B^{2}a-ABb}{A^2+B^2}\end{array}$

To eliminate $x$ and solve for $y$ . So, multiply first equation with B and second equation with − A.

  $\begin{array}{l}\begin{array}{ccc}ABx+B^{2}y& =& BC\\ -ABx+A^{2}y& =& -ABa+A^{2}b\\ y\left(A^2+B^2\right)& =& BC+A^{2}b-ABa\end{array}\\ y=\frac{BC+A^{2}b-ABa}{A^2+B^2}\end{array}$

Thus, the point of intersection is at $Q\left(\frac{AC+B^{2}a-ABb}{A^2+B^2},\frac{BC+A^{2}b-ABa}{A^2+B^2}\right)$ .

c.

To determine

To find: The distance between point $P\left(a,b\right)$ and $Q\left(\frac{AC+B^{2}a-ABb}{A^2+B^2},\frac{BC+A^{2}b-ABa}{A^2+B^2}\right)$ .

Answer to Problem 58E

  $d=\left|\frac{aA+bB-C}{\sqrt{A^2+B^2}}\right|$

Explanation of Solution

Given Information:

  $P\left(a,b\right)$ and $Q\left(\frac{AC+B^{2}a-ABb}{A^2+B^2},\frac{BC+A^{2}b-ABa}{A^2+B^2}\right)$

Calculation:

Using distance formula to find the distance between, $P\left(a,b\right)$ and $Q\left(\frac{AC+B^{2}a-ABb}{A^2+B^2},\frac{BC+A^{2}b-ABa}{A^2+B^2}\right)$

  $\begin{array}{l}Q\left(\frac{AC+B^{2}a-ABb}{A^2+B^2},\frac{BC+A^{2}b-ABa}{A^2+B^2}\right)\\ d=\sqrt{{\left(\frac{AC+B^{2}a-ABb}{A^2+B^2}-a\right)}^2+{\left(\frac{BC+A^{2}b-ABa}{A^2+B^2}-b\right)}^2}\\ d=\sqrt{{\left(\frac{AC+B^{2}a-ABb-a{A}^2-a{B}^2}{A^2+B^2}\right)}^2+{\left(\frac{BC+A^{2}b-ABa-b{A}^2-b{B}^2}{A^2+B^2}\right)}^2}\\ d=\sqrt{{\left(\frac{AC-ABb-a{A}^2}{A^2+B^2}\right)}^2+{\left(\frac{BC-ABa-b{B}^2}{A^2+B^2}\right)}^2}\\ d=\frac{\sqrt{\left(A^2+B^2\right){\left(aA+bB-C\right)}^2}}{A^2+B^2}\\ d=\left|\frac{aA+bB-C}{\sqrt{A^2+B^2}}\right|\end{array}$

Thus, the distance is $d=\left|\frac{aA+bB-C}{\sqrt{A^2+B^2}}\right|$