Multiple choice question:
For what values of $a$ will the lines $ay+3x=4$ and $2y+4x=3$ have no intersection points?
- A) 2
- B) 1.5
- C) 8/3
- D) -0.5
- E) -2
I tried rearranging them into $y=mx+b$ form and solve, but I will end up with a literal answer like this
$$a=\frac{(4-3x)}{y}$$
How do you work this out? I think you need to make the gradient the same but how do you do that?
On
Consider the system
then if $a=\frac32$ LHS become equals but RHS are different thus the system has no solution and we don't have intersection points.
On
They will have no intersection points when and only when the lines are distinct parallel lines. Since the second line is $2y+4x=3$, the lines will be parallel when the first line is $\lambda\times(2y+4x)=\alpha$, for some numbers $\alpha$ and $\lambda$ with $\lambda\neq0$. But the first line is $ay+3x=4$ and therefore $\lambda$ can only be $\frac32$ and in this case the lines are indeed distinct lines.
On
Transforming the two to the form $y=mx+b$ we get the two lines $$ y=-\frac 3ax+\frac4a\\ y=-\frac42x+\frac32 $$ If the lines don't intersect, they must have the same slope. So set the two slopes equal, and solve for $a$.
After we've found our $a$, we must remember to check that there are indeed no solutions, rather than infinitely many. We do this by inserting the value for $a$ that we found, solve the system of equations the regular way, and make sure that we get something like $0=1$ at the end instead of, say, $4=4$. Alternatively, insert that value of $a$ into our $y=mx+b$ form equations above and confirm that we get lines with different constant term.
On
Pedestrian approach:
Subtract $3$× second equation from from $4$×first equation to eliminate $x:$
$4ay-6y= 7;$
$2y(2a- 3) =7;$
$\rightarrow :$
$2a-3 \not =0$ (why?).
On
If the lines do not intersect it means that they are parallel and their slope must be equal. That is:
m1 = m2. Where m1 =`-coff. of x of first eq./coff. Of y of first eq.
& m2= - coff. of x of second eq./ coff. of y of second eq.
i.e.,
-a/3 = -2/4 Solving this we get
a = 3/2.
On
The system is equivalent to $$ \begin{bmatrix} 3&a\\ 4&2 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} =\begin{bmatrix} 4\\ 3 \end{bmatrix}. $$ Let $A$ be the $2\times 2$ matrix above. The lines don't intersect iff $(4, 3)^T$ is not in the span of the columns of $A$. The lines don't intersect iff the columns are linearly dependent (as $(4,3)^T\notin\text{span} \{(3,4)^T$}. Thus the lines don't intersect iff $$ \lambda \begin{bmatrix} a\\ 2 \end{bmatrix} = \begin{bmatrix} 3\\ 4 \end{bmatrix} $$ for some $\lambda\in\mathbb{R}$ whence $\lambda a=3$ i.e. $a=3/2$.
If the slopes (or gradients) of the lines are the same, as @DavidG.Stork puts it, then the lines will be parallel and never meet. Remember that the Cartesian Plane is infinitely long and wide, so if the lines are not parallel, they will eventually meet and thus intersect. Therefore, the lines must be parallel, and so the slopes must also be equal to each other. In both the equations, we have $$y = -\frac{3}{a}x + \frac{4}{a}\,\text{ and }\,y = -2x + \frac{3}{2}$$ so if the slopes are the same, we have $-2 = -\dfrac{3}{a}$ and therefore, the negatives cancel out and $a = \dfrac{3}{2}$. We conclude by stating that the correct answer is $B)$.
This is because $m$ represents the slope in the linear equation, $y = mx + b$.