Is it even possible to build a line having slope of $3$? Could it be a mistake?
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4$\begingroup$ Slopes run from $-\infty$ to $\infty$ (excluded). What makes you think differently ? $\endgroup$– user65203Nov 25, 2016 at 20:45
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$\begingroup$ Looks like there is a direct example of it working, so I'll guess that yes it is possible. $\endgroup$– Kitter CatterNov 25, 2016 at 20:46
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5$\begingroup$ Of course it is. Why do you think that a slope of $3$ wouldn't work? $\endgroup$– user137731Nov 25, 2016 at 20:46
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3$\begingroup$ To be more useful the graph you posted doesn't have the point $(-2,-8)$ plotted it has $(-2,0)$ and $(0,-8)$ plotted. These are different. $\endgroup$– Kitter CatterNov 25, 2016 at 20:48
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1$\begingroup$ My mistake... I built a line based on the coordinates of a point! I didn't pay attention to the problem description and marked separately point x-coordinate as a point and y-coordinate as a point and then joined them together... sorry! $\endgroup$– loewitzNov 25, 2016 at 21:14
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4 Answers
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A slope of $3 = \frac{3}{1}$ means that starting at any point on the line (like say $(-2,-8)$), if you move $3$ units up and $1$ unit right then you'll get another point on the line.
So, yes. It is possible.
My advice is to try playing around with different points and slopes on WolframAlpha to get a feel for it.
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Not a mistake. You seem to have a very basic misunderstanding. The line you drew passes through the TWO points (-2,0) and (0,-8). Its slope is -4. That's just one problem.
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Slope of a line is nothing but the tangent of angle which the line make with $x$-axis or simply speaking $\tan\theta$.
Since value of $\tan\theta$ vary from $\infty$ to $-\infty$ so yes a slope of 3 for a line is possible.
Now, you may have the question that How to construct a line having slope 3? Since you have the point (-2,-8), just plot it on cartesian axes and extend a line from that point such that on moving along line you will move a distance of 3 units on $y$-axis for a distance of 1 unit on $x$-axis.
answered Nov 25, 2016 at 21:19
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$\begingroup$ then writing the equation in the slope y-intercept form, isn't it supposed to look like this: y=3/1x-2=0 ? $\endgroup$– loewitzNov 25, 2016 at 21:28
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$\begingroup$ Yes @loewitz you can use the slope intercept form. The equation will be same as you stated in the comment. But there the zero is invalid. The equation will be y=3/1x-2 $\endgroup$ Nov 25, 2016 at 21:31
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$\begingroup$ When writing the equation in the slope y-intercept form I understand I have to write y=... and this is the beginning of the equation, I also understand I need to write 3/1 because that is the fraction representing the slope... I also understand writing the '-2' part because that is where the line is crossing the y-axis... but why do I have to write the 'x' ??? $\endgroup$– loewitzNov 25, 2016 at 21:47
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$\begingroup$ Whenever you are giving the equation of a line in a plane (in 2-Dimentions) you have to specify or establish a relation between x and y. $\endgroup$ Nov 25, 2016 at 21:49
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$\begingroup$ so, I understand that in the left part of such equation I will have to write Y=... and then the right part of the equation has to contain 'X'? $\endgroup$– loewitzNov 25, 2016 at 21:51
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You should think of a line with its general form of equation.
$ax + by + c = 0$
Then you can represent a vertical line when $b = 0$ The same goes for horizontal ones when $a = 0$
The only constrains come if $a = 0$ and $b = 0$, then it does not represent a line any more.
