Can I have help writing an equation for Line m has no y-intercept, and its x-intercept is (3, 0). Line n has no x-intercept, and its y-intercept is (0, -4). I need to write one for line m and line n
@mathmale
@roadjester
@DanJS
Would it be simply x=3?
I'm confused
Yes, you're right! Using that same logic you can work out line n too.
so simply y=-4 thanks
How so?
a line without one of the axis intercepts, will be parallel to that axis,
any form of slope at all will have all real values for x and y
That's right. This is an example of a question where the lines don't perfectly fit the model y = mx + b. Technically line n does, but in a strange way. We know m is the gradient of the line and b is the y intercept, so... For line n, the y intercept is -4. As the line never crosses the x axis, it must be horizontal (m = 0). If m = 0, mx = 0x = 0. So y = 0 - 4 y = -4 Line m is a bit more tricky. There is NO y intercept! Which means you don't have a "b" value at all! The "b" value isn't 0 (because that means the y intercept is 0, which is different from nothing). So it's best to ignore y = mx + b for this one and just recognise that it's a vertical line. If we think about it logically, x = 3 fits nicely as a description of the line because it's saying "for every point on the line, the x value is 3, regardless of the y value".
yeah, to make a line not cross y-axis, it will be parallel to y axis (vertical)... and also pass through point (3,0), so it has to be located at x=3
Yeah I'm using desmos graphing calculator and put x=3 and y=-4 and it gave me lines so idk what I should put
the equation x=3 or y=4 , can be shown as a single value point on the number line, but when you move to the plane, x=3 is now a line instead of a point x=3 , and y has no restriction
x = 3, all the points (3, y)
same thing for moving to 3 space, - another dimension added lines on the (x,y) plane , turn to planes on 3-space, z has no restriction
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