# Toronto Math Forum

## MAT334-2018F => MAT334--Lectures & Home Assignments => Topic started by: Vedant Shah on September 23, 2018, 05:19:33 PM

Title: Section 1.2 Question 18
Post by: Vedant Shah on September 23, 2018, 05:19:33 PM

Show that two lines $\Re(a+ib)=0$ and $\Re(c+id)=0$ are perpendicular  $\iff \Re(a \bar{c}) = 0$
From section 1.2: Let $a = A+iB$ and $c= C+iD$. Then the lines are $Ax-By+\Re(b)=0$ and $Cx-Dy+\Re(d)=0$
Setting the slope of the first equal to the negative reciprocal of the other I get: $\frac{A}{B} = - \frac{D}{C} \iff AC=-BD$
Finally, $\Re(a \bar{c}) = AC-BD= 2AC$

How do I proceed?

Thanks!

Title: Re: Section 1.2 Question 18
Post by: Victor Ivrii on September 23, 2018, 09:11:34 PM
Consider arguments of two $a,c$
Title: Re: Section 1.2 Question 18
Post by: Min Gyu Woo on September 24, 2018, 02:15:00 PM
Isn't $Re(a\overline{c}) = AC + BD$?
Title: Re: Section 1.2 Question 18
Post by: Hehree Lee on October 08, 2018, 07:26:55 PM
Sorry, a little late, but I personally used another form of the line equation, Re[(m + 1)z + b], to relate the two perpendicular lines and solve the problem.