Find the equation of the line passes through (1, 2) and perpendicular to the line 2y - x + 4 = 0
we have the point (1, 2)
Then the equation for the line is:
y- yA = m (x-xA)
where m is the slope.
But the slope for the line perpendiclaur is:
2y - x + 4 =0
2y = x - 4
==> y = (1/2)x + 2
The slope is (1/2)
But the product of the slopes of two perpedicular lines = -1
==> (1/2)*m = -1
==> m = -2
==> y - 2 = -2 (x -1)
==> y = -2x + 2 + 2
==> y = -2x + 4
The equation of a line perpendicular to another has the slope -1/m.
In the above equation,
2y - x + 4 = 0
y = 1/2 x -2
So the slope of the above line is m = 1 /2
The equation of the line that passes though (1,2) perpendicular to the above line has the slope -2
The equation of a line is
(y - y1) = m(x - x1)
(y - 2) = -2(x - 1)
y = -2x + 2 + 2
y = -2x + 4
A line that is perpendiculato ax+by+c = o is got by swapping the coefficients of x and y and putting a sign to any one of them:
So bx-ay+k = 0.If this passes through a point (x1,y1), then:
b(x-x1) -a(y-y1) = 0
So in the present case, (x1,y1) = (1,2). and ax+by+c = 2y-x+4.
Therefore the perpendicular line to this through (1,2) is:
is: 2x +y+k = 0, as this passes through (1,2) the line becomes 2(x-1)+(y-2) = 0.