# Write the equation of the line that passes through the point (1,2) and the midpoint of the line that passes through (2,3) and (2,-5).Write the equation of the line that passes through the point...

Write the equation of the line that passes through the point (1,2) and the midpoint of the line that passes through (2,3) and (2,-5).

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The line passes through :

A ( 1, 2)

and B(xB, yB) where B is midpoint of (2,3) and (2,-5)

Let us find B:

xB = (2+2)/2 = 2

yB = 3+-5/2 = -2/2 = -1

Then the point B (2,-1)

Now we have A and B passes through the line:

==> y-yA = m(x-xA)

m = (yB-yA)/(xB-xA) = (-1-2)/(2-1) = -3

==> y-2 = -3(x-1)

==> y-2 = -3x + 3

**==> y= -3x + 5**

The mid point of (2,3) and (2,-5) is [(2+2)/2, (3-5)/2] or (2, -1)

The other point we are given is (1, 2)

So we need the equation of the line that passes through (2, -1) and (1, 2) .

We know that the equation of the line through ( x1, y1) and (x2, y2) is given by: y-y1 = [(y2-y1)/(x2-x1)]*(x-x1)

Here, substituting the values we get:

y-(-1)= [(2- (-1))/(1- 2)]*(x- 2)

=> y+1 = [(2+1)/-1]* (x-2)

=> y+1 = -3 (x-2)

=> y+1 = -3x + 6

=> 3x + y - 5 = 0

**So the required equation is 3x + y - 5 = 0**

The equation of the line passing through (x1,y1) and (x2,y2) is given by"

y-y1 = {(y2-y1)/(x2-x1)} (x-x1).

Therefore (x1 ,y1) = (2,3) and (x2,y2) = (2,-5) . Here x2-x2 = 0. So , we can write the equation:

(x-x1) = ((X2-x1)/(y2-y1)}(y-y1)

x-2 = {(2-2)/(-5-3)(y--3)

x-2 = 0*(y-3)/-8 = 0

Sox-2 = 0 Or x= 0 is the equation of the line.

We'll determine the coordinates of the midpoint of the line that passes through the points (2,3) and (2,-5).

xM=(x2+x1)/2=(2+2)/2=2

yM= (y2+y1)/2=(3-5)/2=-1

In order to write the equation of the line that passes through the points N (1,2) and M(2,-1), we have to determine the slope of this line mMN = (yM-yN)/(xM-xN)

mMN = (-1-2)/(2-1)

mMN = -3

The equation of the line that passes through the points MN IS:

y-yN=mMN*(x-xN)

y-2=(-3)*(x-1)

We'll remove the brackets:

y-2+3x-3=0

**y+3x-5=0**