Question 29359: I have not taken an Algebra class in 30 years,.. This has stumped me, even after referring to the text…
Determine the equation of the following line. Write the answer in standard form using only integers as coefficients.:
…..The line through (-2, 4) that is perpendicular to the x-axis.
thanks in advance me
Found 2 solutions by atif.muhammad, sdmmadam@yahoo.com:
Answer by atif.muhammad(135)   (Show Source):
You can put this solution on YOUR website! It is perpendicular to the x-axis
This means that it has a gradient of 0, it is a horizontal line.
Formula to find equation of line:
(y-y1) = m(x-x1) m stands for gradient
gradient = 0
(-2,4) --> Substitute these values into the formula, -2 is y1, 4 is x1
y-4 = 0(x--2)
y-4 = 0(x+2)
y-4 = 0
y=4 --> This is the equation of the line.
Answer by sdmmadam@yahoo.com(530) (Show Source):
You can put this solution on YOUR website! Determine the equation of the following line. Write the answer in standard form using only integers as coefficients.:
…..The line through (-2, 4) that is perpendicular to the x-axis.
The line through (-2, 4) that is perpendicular to the x-axis.
Any line perpendicular to the x-axis is parallel to the y-axis
And any line parallel to the y -axis has its equation by
x= k ----(1)
Now P(-2,4) is a point on this line
Therefore putting x=-2 in (1)
(-2) =k
That is k=-2 (*)
Putting (*) in (1)
the required equation is
x=-2
which is a line parallel to the y-axis at a distance 2 units to the left of it
Note: In the above because you do not find any y in the equation (1) you did n't have the opportunity to supply y=4 in the equation to the line. This matter should not leave you perplexed. If that 4 is still disturbing,draw the co-ordinate axes of reference,draw a line parallel to the y-axis at a distance 2 units to the left of the origin.Mark the point P on this vertical line by going 4 units above the x-axis. You reach the point P(-2,4)
Mark any other point or points on this vertical line. all of them will have their x-coordinate given by (-2). Hence the equation to that line is identified by x=-2
Therefore the general rule of thumb is :
Any line parallel to the x-axis will have its equation given by
y= c
where c is the distance between our line and the x-axis
If c >0 that is if c is positive, then the line is above the x-axis
If C <0 that is if c is negative,then the line is below the x-axis
Any line parallel to the y-axis will have its equation given by
x= k
where k is the distance between our line and the y-axis
If k >0 that is if k is positive, then the line is to the right of the y-axis
If k <0 that is if k is negative,then the line is to the left of the y-axis
Note: If you had thought along the lines of slope and one point form of equation to the line there is no harm!
Any line perpendicular to the x-axis makes an angle 90 degrees
with the x-axis and has slope = tan(90)= infinity = (something)/0
The slope and one point form of equation to a line is
(y-y1)/(x-x1) = slope where P(x1,y1) is the given fixed point
And here P(x1,y1)=P(-2,4)
Therefore applying the above equation
(y-4)/([x-(-2)] =(something)/0
which implies the dr = 0
Therefore [x-(-2)] =0
x+2 = 0
That is x= -2
Note:(If a fraction is zero,then the nr =0
and if a fraction is infinity then the dr = 0)
|
|
|
