To find the coordinates of the point which divides in a given ratio (m_{1} : m_{2}) the line joining two given points (x_{1}, y_{1}) and (x_{2}, y_{2}).

Let P_{1} be the point (x_{1}, y_{1}) , P_{2} be the point (x_{2}, y_{2}) and P be the required point, so that we have

P_{1}P: PP_{2} :: m_{1} : m_{2}

Let P be the point (x, y) so that if P_{1}m_{1}, PM and P_{2}m_{2} be drawn parallel to the axis of y to meet the axis of x in m_{1} , M and m_{2}, we have Om_{1} =x_{1}, m_{1}P_{1} = y_{1}, OM = x, MP= y, Om_{2} = x_{2}, and m_{2}P_{2} = y_{2}

Draw P_{1}R_{1} and PR_{2}, Parallel to OX, to meet MP and m_{2}P_{2} in R_{1} and R_{2} respectively.

Then,

P_{1}R_{1} = m_{1}M =OM-Om_{1} = x-x_{1},

PR_{2}= Mm_{2}=Om_{2}-OM=x_{2}-x,

R_{1}P = MP – m_{1}P_{1}= y-y_{1},

R_{2}P_{2} = m_{2}P_{2} –MP =y_{2}y.

From the similar triangles P_{1}R_{1}P and PR_{2}P_{2}, we have

m_{1}/m_{2}=P_{1}P/PP_{2}=P_{1}R_{1}/PR_{2}=(x-x_{1})/(x_{2}-x)

m_{1}(x_{2}-x)=m_{2}(x-x_{1})

x= (m_{1}x_{2}+m_{2}x_{1})/(m_{1}+m_{2})

again, m_{1}/m_{2}=P_{1}P/PP_{2}=R_{1}P/R_{2}P_{2}=(y-y_{1})/9y_{2}-y’)

So that, m_{1}(y_{2}-y)=m_{2}(y-y_{1})

And hence, y=(m_{1}y_{2}+m_{2}y_{1})/(m_{1}+m_{2})

The coordinates of the point which divides P_{1}P_{2} internally in the given ratio m_{1}:m_{2} are therefore,

X = (m_{1}x_{2}+m_{2}x_{1})/(m_{1}+m_{2})

And Y = (m_{1}y_{2}+m_{2}y_{1})/(m_{1}+m_{2})

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 (x _{1}, y_{1}) to (x_{2}, y_{2}) are : (x_{1}+x_{2})/2 and y_{1} +y_{2})/2.*