# The point (1;2) is on the line that is perpendicular to the line x+y-1=0. Determine the equation of the perpendicular line?

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Given the point (1,2) is on the line.

Then we know that:

y-y1 = m(x-x1)

==> y -2 = m(x-1)

Now we need to find the slope m.

But we are given the equation of the line parallel to the equation.

x+y-1 = 0

We will rewrite into the slope form.

==> y= -x + 1

Then the slope is -1.

Since the lines are perpendicular, then the product of the slopes is -1.

==> -1 * m = -1

Then m = 1

==> y-2 = 1 (x-1)

==> y= x -1 + 2

**==> y= x +1 **

**==> x -y +1 = 0**

Any line perpendicular to a line ax+by +c = 0 is of the form bx-ay+k = 0. Further if this line is passing through a particular point (x1,y1) , then b(x-x1) - a(y-y1) = 0.

So the line perpendicular to x-y-1 = 0 passing through (1,2) is the line (x-1)- (y-2) = 0.

=> the line is x-y -1+2 = 0.

=> x-y +1 = 0 is the line passing through (1,2) and perpendicular to x+y -1 = 0.

We know that 2 lines are perpendicular if the product of their slopes is -1.

We'll put the equation of the given line in the point slope form:

y = mx+n

For this reason, we'll keep y to the left side and we'll move the rest of terms to the right side:

y = -x + 1

Comparing, we'll get the slope of this line: m1 = -1

The slope of perpendicular line is m2 = -1/m1

m2 = -1/-1

m2 = 1

The equation of the perpendicular line, that has the slope m2 = 1 and it passes through the point (1;2) is:

y - 2 = 1*(x - 1)

y - 2 = x - 1

x - y - 1 + 2 = 0

x - y + 1 = 0

**The equation of the perpendicular line is:**

**x - y + 1 = 0**