We are looking for pairs of lines with the same slope. We can count the rise and run to determine the slope.

Line a and line d both have a slope of m_a=3=m_d, so they are parallel.

Lines b and c have similar slopes, but are not actually parallel.

We are looking for pairs of lines with slopes that are opposite reciprocals, so their slopes should have a product of -1.

Looking at the coordinate plane, the pairs that look like they might be perpendicular are a and b, a and c, and e and f.

From the coordinate plane, we can find that:

m_a=3,\,m_b=-\dfrac{1}{3},\,m_c=-\dfrac{1}{4},\,m_e=\dfrac{1}{2},\,m_f=-2

Line a and line b have slopes with a product of 3 \left( -\dfrac{1}{3}\right) =-1, so they are perpendicular.

Line e and line f have slopes with a product of \dfrac{1}{2} \left( -2\right) =-1, so they are perpendicular.

Note that we didn't specify anything about line d. This is because in part (a) we determined that lines a and d are parallel, and so any line that is perpendicular to a is also perpendicular to d, by the perpendicular transversal theorem.

In particular, that means that line b is perpendicular to line d.

To find the y-intercept, we can either substitute x=0 or rearrange to slope-intercept form.

Using the strategy of setting x=0,

The y-intercept is \left(0,2\right).

For the new line to be perpendicular to the given line, its slope must be the opposite reciprocal of the slope of the given line.

Rearranging the equation of the given line to slope-intercept form gives us:

The slope of the given line is m=\dfrac{4}{3}.

The slope of a perpendicular line is m_{\perp}=-\dfrac{3}{4}

We know from the previous part that the y-intercept of our new line will be \left( 0,2 \right).

The equation of our new line in slope-intercept form is y=-\dfrac{3}{4}x+2, which we now want to convert to standard form.

The equation of our new line is 3x+4y=8.

Converting the equation of the given line to slope-intecept form confirmed our answer to part (a).

We can do a quick sketch of the normal to help:

Since the mirror is a horizontal line, the normal must be a vertical line if it is to be perpendicular. This means it will be of the form x=a.

Since it goes through the point where the laser hits the mirror, \left(4,0\right), the equation of the normal will be x=4.

Since the angles are congruent, know that the angle formed between the normal and the reflection will also be 45 \degree.

We can label this on our diagram:

Using the angle addition postulate, we can show that the angle formed between the laser and its reflection will be 45 \degree+45 \degree=90\degree.

This means that the laser beam path and the reflection path are perpendicular, so their slopes will be negative reciprocals. Since the slope of the laser beam is -1, this means the slope of the reflection will be 1.

The reflection starts at the point \left(4,0\right). Putting all of this together we get:

The equation of the reflection is y=x-4.

We did not need to be told that the angle formed between the laser beam and the normal was 45 \degree. Since the slope of the line was -1, we could form a right isosceles triangle with legs of length 4, so the angle must be 45 \degree.