Write the point-slope form of the equation of a line with a slope of 0 0 0 that passes through ( 2 , − 4 ) \left(2,-4\right) ( 2 , − 4 ) . Then graph the equation .

Hey, everyone. So let's get started with this practice problem here. So we have to write the point slope form of an equation of a line, and we're told some information here. With the the slope is 0, and it passes through 2 comma 4. So in other words, the m, the slope, is 0, and the point that it passes through, remember, which is just x one y one, is equal to 2 comma negative 4. We take all this information. We just pop it into the point slope form over here, and then we're just gonna basically solve and sort of simplify, and that'll be the equation of our line. So if I have y minus y one equals m parenthesisx minusx1, then I just have to replace, my y one, my m, and my x one into this equation. So my what my y one ends up becoming over here is if you look at this point, the y one is just the negative 4. So I'm just gonna replace this with negative 4. The m in this case, remember the slope is just 0 for this problem. So I had to sort of replace this with a 0, and then this just becomes x minus 2 because that's my x my x one. Now we have just to simplify this, and if you look at the right side, what happens here is that normally we've had slopes that weren't 0 and, you know, you just write like a 2 5ths or a 2 or something like that on the outside, but here it's 0. So what ends up happening is 0 times anything is always gonna be 0 on the right side. So, basically, what you end up with in this equation is that y+4 is equal to 0. It's almost like the x term just goes away because it gets multiplied by 0. So what's going on here? Well, if you take the last step and you just move the 4 over to the other side, what you'll see is that this just becomes y equals negative 4. So we started off with the point slope form of this line over here, and basically what it worked out to be is just the equation y equals some number. And if you remember, this is an equation of a horizontal line. A horizontal line is basically just where you have a y value that's constant, and so you have no x in that expression over there. Just just a horizontal line, and it goes through the point, 0 comma negative 4. Basically, this always remains as 0 as negative 4 as the y value. Alright? So we went from the point slope form of a line, and we, basically, we simplified this to be a perfectly horizontal line, which we've seen before, and we know how to graph that. Alright? So pretty cool. Hopefully, that made sense. Thanks for watching, and I'll see you in the next one.

2. Graphs of Equations

Lines

College Algebra

2. Graphs of Equations

Lines

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Multiple Choice

Write the point-slope form of the equation of a line with a slope of $0$ that passes through $$\left$(2,-4$\right$)$ . Then graph the equation.

$y+4=x-2$

$y+4=x$

$y+4=0$

$y=0$

Verified step by step guidance

Identify the point-slope form of a line equation, which is given by: $ y - y_1 = m(x - x_1) $, where $ m $ is the slope and $ (x_1, y_1) $ is a point on the line.

Substitute the given slope $ m = 0 $ and the point $ (2, -4) $ into the point-slope form equation. This gives: $ y - (-4) = 0(x - 2) $.

Simplify the equation: $ y + 4 = 0 \cdot (x - 2) $. Since the slope is zero, the term $ 0(x - 2) $ becomes zero.

Further simplify the equation to: $ y + 4 = 0 $.

To graph the equation $ y + 4 = 0 $, solve for $ y $ to get $ y = -4 $. This represents a horizontal line passing through $ y = -4 $ on the graph.

5:46m

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Write the point-slope form of the equation of a line with a slope of $0$ that passes through $\(\left\)(2,-4\(\right\))$ . Then graph the equation.