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?
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