Step-by-step explanation:
So it's important to define what parallel is. If one line is parallel to another line, it will never intersect the other line. This means the two lines must be changing by the same amount simultaneously. In other words, they have the same slope. It's also important to remember that same slope doesn't mean they're parallel. You also have to look at the y-intercept. If they have the same y-intercept, and the same slope, then they're the same line. So two lines are only parallel, if the slopes are the same and they have different y-intercepts.
So now that you understand what's parallel let's look at some of the problems
1. So this is kind of tricky, because as I mentioned before, parallel lines have the same slope, but in this case a slope cannot be defined. For two vertical lines to be parallel, they just have to have different a different x-constant. Because x is what's constant, and if they're different, then the two lines will never intersect. This gives us the general formula for a vertical parallel line: [tex]x=a\text{ where a} \ne3[/tex] (this is specific to this problem, since the constant in the graph is x=3). Since the parallel line must pass through (-5, -4) this means that the constant for x, must be -5, because the x is constant, if it equals anything else, it will never equal -5. So the parallel line in this case will be: [tex]x=-5[/tex]
2. So the equation is already in slope-intercept form, so we don't have to solve for the slope. The slope in this case is 8, which can be determined by looking at the formula. This gives us the general formula for the parallel line: [tex]y=8x+b\text{ where b}\ne 4[/tex]. b cannot equal 4, because if it did, then we would have two lines that are the same, not parallel, otherwise b can be anything else. Since we have a point that the parallel line goes through, we can plug this in as (x, y) to solve for b: [tex]21=8(4)+b \implies21=32+b\implies -11=b[/tex]. So now this gives us the complete equation: [tex]y=8x-11[/tex]
3. The image is a bit low quality, although from what I can see the linear equation intersects the origin (0, 0) and (4, -1). This gives us the equation for the slope -1/4 (rise/run). which gives is the slope -1/4. So this gives us the general equation: [tex]y=-\frac{1}{4}x+b \text{ where b} \ne0[/tex]. b cannot equal 0, since it would then have the same y-intercept, meaning the two lines would be the same. In this case we also get some coordinates which we can plug into the equation to solve for b: [tex]-5=-\frac{1}{4}(12)+b\implies -5=-3+b\implies-2=b[/tex]. This gives us the complete formula: [tex]y=-\frac{1}{4}x-2[/tex]
4. This is similar to the vertical line, as there is a constant value, but in this case a slope can be defined, and that slope is 0; it's 0 for all horizontal lines. So we have the general equation: [tex]y=0x+b \text{ where b}\ne4[/tex]. This can be simplified to: [tex]y=b \text{ where b} \ne4[/tex]. Since the line passes through (13, -7), that means b must equal -7, because the y does not depend on x. It's a constant value. So you have the equation: [tex]y=-7[/tex]
5. [tex]y-6=\frac{3}{5}(x+5)[/tex]. So in this case we have the point-slope form. We don't need to convert it to the slope-intercept form, since all we need to form the parallel line is the slope, sort of. You technically do need the y-intercept, so you know what b cannot equal, so that you don't have the same line, but I'm assuming, the coordinates they're giving you, do not pass through the original line, but rather a parallel line. So we have the general equation: [tex]y=\frac{3}{5}x+b[/tex]. Plugging the given coordinates give you the equation: [tex]8=\frac{3}{5}(0)+b \implies8=b[/tex]. You didn't really need to do this, since if you noticed the coordinates have x=0, which means the y-value by definition is what b is equal to, since b is the y-intercept. So this gives you the equation: [tex]y=\frac{3}{5}x+8[/tex]
