# Coordinates of the point which divides in a given ratio

To find the coordinates of the point which divides in a given ratio (m1 : m2) the line joining two given points (x1, y1) and (x2, y2).

Let P1 be the point (x1, y1) , P2 be the point (x2, y2) and P be the required point, so that we have
P1P: PP2 :: m1 : m2
Let P be the point (x, y) so that if P1m1, PM and P2m2 be drawn parallel to the axis of y to meet the axis of x in m1 , M and m2, we have Om1 =x1, m1P1 = y1, OM = x, MP= y, Om2 = x2, and m2P2 = y2
Draw P1R1 and PR2, Parallel to OX, to meet MP and m2P2 in R1 and R2 respectively.
Then,

P1R1 = m1M =OM-Om1 = x-x1,
PR2= Mm2=Om2-OM=x2-x,
R1P = MP – m1P1= y-y1,
R2P2 = m2P2 –MP =y2y.
From the similar triangles P1R1P and PR2P2, we have

m1/m2=P1P/PP2=P1R1/PR2=(x-x1)/(x2-x)
m1(x2-x)=m2(x-x1)
x= (m1x2+m2x1)/(m1+m2)
again, m1/m2=P1P/PP2=R1P/R2P2=(y-y1)/9y2-y’)
So that, m1(y2-y)=m2(y-y1)
And hence, y=(m1y2+m2y1)/(m1+m2)

The coordinates of the point which divides P1P2 internally in the given ratio m1:m2 are therefore,
X = (m1x2+m2x1)/(m1+m2)
And Y = (m1y2+m2y1)/(m1+m2)

The proof of this statement is similar to that of the preceding article and is left as an exercise for the student.
Note:The coordinates of the middle point of the line joining (x1, y1) to (x2, y2) are : (x1+x2)/2 and y1 +y2)/2.