# If the straight lines $3x − y = −2$ and $ax + 2y = 3$ are parallel, then $a$ equals?

i. If the straight lines $3x − y = −2$ and $ax + 2y = 3$ are parallel, then $a$ is equal to what?

ii. If the straight lines $5x + y − 3 = 0$ and $bx − y − 2 = 0$ are perpendicular, then $b$ is equal to what?

I am confused with these. I'm not even sure where to start? Any advice/help?

Thank you.

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What does being parallel (or perpendicular) mean in terms of the slope? –  John Habert Jan 29 '14 at 1:26
The product of two gradients equals -1? –  Orbit Jan 29 '14 at 1:27
That is true for the perpendicular case. So find the gradients of each line, multiply and set equal to $-1$. Solve for $b$. What about the parallel case? –  John Habert Jan 29 '14 at 1:28
For parallel the gradients are the same –  Orbit Jan 29 '14 at 1:31
So find both gradients and set them equal and that allows you to solve for $a$. –  John Habert Jan 29 '14 at 1:31

i. If the straight lines $3x − y = −2$ and $ax + 2y = 3$ are parallel, then $a$ is equal to what?

Hint:
First, and for the sake of simplicity, write $3x-y=-2$ as: $$y=\color{blue}3x+2\tag{1}$$ and $ax+2y=3$ as $$y=\color{blue}{-\dfrac a2}x+\dfrac32 \tag{ 2}.$$ We can now clearly determine the slopes (gradients) of the two lines from just looking at the coefficients of $x$ in $\text{(1)}$ and $\text{(2)}$ which turns out to be $m=3$ and $m'={-a/2}$.

Next, use this property:

Let the line $(\rm D)$ be defined by the equation: $y=mx+b$ and the line $(\rm \Delta)$ be defined by the equation: $y=m'x+b'$.

If $(\rm D)$ is parallel to $(\Delta)$ then $m= m'$ and the opposite is true.

So you'll end up with the equation: $$3=-\dfrac a2$$ Solve it, and here is the value of $a$.

ii. If the straight lines $5x + y − 3 = 0$ and $bx − y − 2 = 0$ are perpendicular, then $b$ is equal to what?

Hint:
Again, for the sake of simplicity we will reduce $5x+y-3=0$ to $$y=\color{blue}{-5}x+3\tag{3}$$ and the equation $bx-y-2=0$ to $$y=\color{blue}{b}x-2\tag{4}$$

So the slopes of the two lines are $m=-5$ and $m'=b$ respectively.

Now that we have the slopes (gradients), will use this property:

Let the line $(\rm D)$ be defined by the equation: $y=mx+b$ and the line $(\rm \Delta)$ be defined by the equation: $y=m'x+b'$.

If $(\rm D)$ is perpendicular to $(\Delta)$ then $m\cdot m'=-1$ and the opposite is true.

Therefore we have: $$-5\cdot b=-1.$$

Now, solve the equation and here is your value of $b$!

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In part (i), the gradients are $3$ and $-\frac{a}{2}$, respectively. Since they are parallel, you can set them equal and solve. This yields

$-\dfrac{a}{2} = 3\\ a = -6$

In part (ii), the gradients are $-5$ and $b$, respectively. Since they are perpendicular, their product is $-1$. This yields

$-5*b = -1 \\ b = \dfrac{-1}{-5} = \dfrac{1}{5}$

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Thank you that explains it. –  Orbit Jan 29 '14 at 1:58

The key to this problem deals with the slopes of perpendicular and parallel lines.

Hint: one of them has something to do with a "negative reciprocal," or a product being equal to $-1$. That may help you in looking through your notes.

Use algebra to convert the equations to slope-intercept ($y=mx+b$) form, then use the relationships between slopes to find the variables you seek.

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