# 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 or parallel to the x-axis, that means that the slope (m)=0

We hav the point (6,-1) pass through the line:

y-y1= m(x-x1)

y+1=0

y=-1

The line is y=-1

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**

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.

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