# Find the equation of the line passes through (1, 2) and perpendicular to the line 2y - x + 4 = 0

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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.

2x+y-4 =0