$d_1:ax+2y+4=0$; $d_2:3x-4y+12=0$; $|d_1d_2|=?$

Question

$d_1:ax+2y+4=0$

$d_2:3x-4y+12=0$

$d_1$ // $d_2$

$|d_1d_2|=?$

This is a slightly modified version of a university entrance examination problem set up by the experts at this specific university itself.

I don't know what is meant by $|d_1d_2|$. I searched the internet for what could have been meant by this expression, but I couldn't find it. When taken as the distance between two parallel lines, the answer can be found to be $4$.

I want to object to the problem for the ambiguity on what is meant by $|d_1d_2|$. During the rush of the exam, I thought the problem asked for the absolute value of the multiplication of the two lines, for whatever it would be.

Is this expression ,$|d_1d_2|$, algebraically correct and does it mean the distance between two parallel lines?

Answer 1

Since $d_1||d_2$ we have $-{a\over 2} = {3\over 4}$ so $a= -{3\over 2}$. Now take any point $T(x_0,y_0)$ from one line, say $d_1:\;\; -3x+4y+4=0$ and calcluate it distance from $d_2$, say by formula: $$ {|ax_0+by_0+c|\over \sqrt{a^2+b^2}}$$

Answer 2

You can safely skip the computations in 1. and go directely to 2. if you are only interested in the notation.

1. Distance between $d_1$ and $d_2$. Two straight lines represented by the equations $$Ax+Bx+C=0\qquad\text{ and } \qquad A'x+B'x+C'=0$$ are parallel if and only if $$ AB'=A'B .$$ This means that $$d_1:ax+2y+4=0, \qquad \text{ with }\qquad A=a, B=2, C=4$$ and $$d_2:3x-4y+12=0\qquad \text{ with }\qquad A'=3, B'=-4, C'=12$$ are parallel if and only if $$ a(-4)=3(2)\iff a=-\frac{3}{2} .$$ Now, let $d$ be a straight line with equation $$d:\tilde{A}x+\tilde{B}y+\tilde{C}=0.$$ This line $d$ is perpendicular to e.g. $d_1$ if and only if $$A\tilde{A}+B\tilde{B}=0\iff \tilde{A}=-\frac{B\tilde{B}}{A} ,\qquad \text { with } A=a=-\frac{3}{2}, B=2,$$ which means that $$d: \tilde{A}x+\tilde{B}y+\tilde{C}=0\iff -\frac{B\tilde{B}}{A}x+\tilde{B}y+\tilde{C}=0 \iff Bx-Ay-\frac{A\tilde{C}}{\tilde{B}}=0.$$ Hence $d$ is represented by $$d: 2x+\dfrac{3}{2}y+\frac{3}{4}\tilde{C}=0.$$ For $\tilde{C}=0 $, we get a line $d'$ parallel to $d$: $$d': 2x+\dfrac{3}{2}y=0.$$ If we compute the coordinates of the intersection points of $d_1$ and $d_2$ with $d'$, we find that $$ \left\{ \begin{align} -\frac{3}{2}x+2y+4&=0 \\[2ex] 2x+\frac{3}{2}y&=0 \end{align} \right. \iff \left\{ \begin{array}{l} x=\dfrac{24}{25} \\[2ex] y=-\dfrac{32}{25} \end{array} \right. $$ and $$ \left\{ \begin{align} 3x-4y+12&=0 \\[2ex] 2x+\frac{3}{2}y&=0 \end{align} \right. \iff \left\{ \begin{array}{l} x=-\dfrac{36}{25} \\[2ex] y=\dfrac{48}{25}. \end{array} \right. $$ As you claimed the distance between $d_1$ and $d_2$ is indeed $4$, because $$\sqrt{\left( \frac{24}{25}+\frac{36}{25}\right) ^{2}+\left( -\frac{32}{25}- \frac{48}{25}\right) ^{2}}=4.$$

   2. Notation

      • As for the notation of the length of the line segment $AB$, that is the distance between two points $A$ and $B$ is normally written $\overline{AB}$. However, $\overline{AB}$ may also denote the line segment itself (here) instead of the length of the line segment, which is then denoted by $AB$. Another notation for the line segment is $[A,B]$ (here). The symbol $|\ldots|$ may denote Euclidean norm or Euclidean length or magnitude (here)
      • Concerning the notation $|d_1d_2|$, I've never seen it before, but from the context, the answer to your last question "does it mean the distance between [the given] parallel lines?" seems reasonable to assume so.