Find the equation of the line through the point (6 , -1) and is perpendicular to the y axis.
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The line is perpendicular to the y-axis means that it is horizontal...
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Let's put the problem in this way: if the line is perpendicular to y-axis, then it is parallel to x-axis.
The slopes of 2 parallel lines are equal:
m1=m2
But the slope of the x-axis is m1=0, so m2=0, also.
Now, let's write the equation of a line which passes through a given point A(6 , -1) and it has a known slope, m2=0:
y - yA = m2(x - xA)
y - (-1) = 0*(x - 6)
y + 1 = 0
y = -1
The equation of the line which is perpendicular to y-axis is:
y = -1
Any point on a line that is perpendicular to y-axis has the same y coordinate:
Therefore general equation for a line perpendicular to y-axis is:
y = c
Where c is a constant.
The line specified in the question passes through the point (6, -1). That is for this point y = -1.
Therefore for a line perpendicular to y-axis and passing through the given point:
c = -1
And equation of this line is:
y = -1
Let us assume the line to be of the standard form: y = mx+k., where m is the slope of the line and k is a constant.
Now m and k we decid by the two given conditions: The slope m is zero for any line perpendicular to y axis Orparallel to X axis.
So the line is y = k.
The second condition is the line passes thruogh (6,-1). So the coordinates of this point should satisfy the equation y= k. Or y = 0*x+k. So putting x=6 and y = -6 in y = 0x +k, we get:
-1 = 0*(6)+k. This determines k = -1. So m = 0 . k =-1. Substituting these values in the assumed equation y = mx+k, we get:
y = -1. Or
y+1 = 0 is the required eqation of the line.
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