Which of the following is the equation of the line in the xy-plane tha

How can I tackle this question? The answer is B but I do not understand why?
If the equation of a line is in the form y = mx + b. How could y = 3 have a slope of 0? Moreover, why can't statement I be the answer?

Which of the following is the equation of a line in the xy-plane that has slope 0?

I. x=2
II. y=3
III. x+y=0

A) I only
B) II only
C) III only
D) I and II
E) II and III

So I'm a little confused. I thought for Statement I and II when the equation of a line is like that it always means the slope is 0.

For Statement I we know that the X coordinate is always 2 so one method is to plug in coordinates for Y and then flind the slope so (2,10) and (2,-3) so we get -3-10/2-2 =-13/0=undefined

Then for statement II same idea. Y is always 3 so we can do (2,3) and (5,3). Slope will equal 0.

Statement III confuses me. I guess I always thought x=3 the slope was 0? Can someone clarify?

The equation of a line is y = mx + c, where m is the slope and c is the y intercept.
Slope 0 implies that the line is parallel to the x axis. which means that y intercept should be constant and x intercept 0 i.e the line will never intersect the X axis or is the x axis itself
clearly, y = (m=0) (x = 0) + c
i.e y = c will be the equation of such a line. In this case y = 3.
In general y = n ( where n is -infinity to infinity) is the equation of a line with slope 0.
Hope this is clear.

Points related to slope:


1. If a line is vertical, the slope is not defined, line is parallel to Y-axis and crosses quadrant I and IV, if the X
intersect is positive and quadrant II and III, if the X intersect is negative. Equation of such line is ,
where a is x-intercept.

In our case x = 2. undefined slope.

2. If a line is horizontal it has a slope of 0 , is parallel to X-axis and crosses quadrant I and II if the Y intersect is
positive OR quadrants III and IV, if the Y intersect is negative. Equation of such line is y=b, where b is y intersect.

In our case y = 3. Slope is 0.

3. The general form of the equation is ax+by+c =0 .

In our case y = -x, we have negative slope.

Hope this helps.

msk0657: Couldn't quite understand the concept in point no:3. General form of the equation is ax+by+c=0.

If y=-x, then substituting the same is the equation y=mx+b, would give -1=m+b, m=-1-b. But then without knowing value of 'b', 'm' cannot be determined. Can you explain.?

Hi Narendran,

1. General form.
The general form of the equation of a straight line is ax+by+c = 0.
Where a,b and c are arbitrary constants. This form includes all other forms as special cases. For an equation in this form the slope is -a/b and the y intercept is -c/b.

2. Point-intercept form.
y = mx+b.
Where: m is the slope of the line; b is the y-intercept of the line; x is the independent variable of the function y .


We know the standard equation form is ax+by+c=0 , which you mentioned correctly. We can write like this y = (-a/b)x + (-c/a) , here y intercept is (-c/a) and slope is (-a/b).

We are given only y+x = 0 and let's write this one in the above converted form i.e. => y = (-1)x + 0 , we are not given any b value i.e. y-intercept value and if we compare equation with the standard one then we can see slope is -1 and it is -ve value.

Hope this is clear...Please let me know if this is not clear.

msk0657: Totally understood your explanation. Thank you!

Can you pls explain the slopes in the 2nd quadrant and how they are derived?

Hi, I don't get the sentence marked in red above. How does substituting m=0 give y=3? It should give y=C. What am I missing here?

x=2 is a vertical line, slope is undefined
y=3 is a horiztontal line, slope = 0
x+y=0 is the same as y=-x, negative slope.

Thus, II is correct--> B

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