Question

Find the coordinates of the points which divide the line segment joining \[A( - 2,2)\] and \[B(2,8)\] into four equal parts.

Answer 1

Hint: Here, a given coordinate \[A\] and \[B\] a line segment which divides the coordinates into four equal parts. Here the four equal parts are along with the coordinates.
The line is divided into equal parts; each part is equal to the other part.
Assuming the part is equal to some constant, hence derive the sum.
Finally we get the answer

Complete step-by-step answer:
Let the points that divide \[AB\] into \[4\] equal parts be \[{P_1},{P_2},{P_3}\]
          

We know that, the line segment joining \[A\] and \[B\] into four equal parts
\[A{P_1} = {P_1}{P_2} = {P_2}{P_3} = {P_3}B\]
Assuming
\[A{P_1} = {P_1}{P_2} = {P_2}{P_3} = {P_3}B = k\]
Hence, the section formula tells the coordinates of the point which divides a given line segment into two parts such that their lengths are in the ratio \[m:n\].
Taking \[A{P_2}\] by \[{P_2}B\] is equal to \[\dfrac{{A{P_1} + {P_1}{P_2}}}{{{P_2}{P_3} + {P_3}B}}\]
It is of the form,
\[\dfrac{{A{P_2}}}{{{P_2}B}} = \dfrac{{A{P_1} + {P_1}{P_2}}}{{{P_2}{P_3} + {P_3}B}}\]
Substituting the above values we get,
\[\dfrac{{A{P_2}}}{{{P_2}B}} = \dfrac{{k + k}}{{k + k}}\]
Adding the terms we get,
\[\dfrac{{A{P_2}}}{{{P_2}B}} = \dfrac{{2k}}{{2k}}\]
Cancelling the terms we get,
\[\dfrac{{A{P_2}}}{{{P_2}B}} = \dfrac{1}{1}\]
We can write it in the form of ratio,
\[A{P_2}:{P_2}B = 1:1\]
Hence point \[{P_2}\] divides \[AB\] into equal parts \[A{P_2}\] and \[{P_2}B\]
Hence the coordinate of \[{P_2}\] are \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)\]
Here \[A( - 2,2) = A({x_1},{x_2})\]and \[B(2,8) = B\left( {{y_1},{y_2}} \right)\]we get,
\[ = \left( {\dfrac{{ - 2 + 2}}{2},\dfrac{{2 + 8}}{2}} \right)\]
= \[\left( {\dfrac{0}{2},\dfrac{{10}}{2}} \right)\]
= \[(0,5)\]
So, \[{P_2}(0,5)\]
Similarly, to find \[{P_1}\]
Take \[A{P_1}\] by \[{P_1}{P_2}\] is equal to \[\dfrac{k}{k}\]
\[\dfrac{{A{P_1}}}{{{P_1}{P_2}}} = \dfrac{k}{k}\]
Cancelling the terms we get,
\[\dfrac{{A{P_1}}}{{{P_1}{P_2}}} = \dfrac{1}{1}\]
\[A{P_1}:{P_1}{P_2} = 1:1\]
Hence point \[{P_1}\] divides \[A{P_2}\] into two equal parts
Hence the coordinates of \[{P_1}\] are
\[ = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)\]
Here \[A( - 2,2) = A({x_1},{x_2})\] and \[{P_2}(0,5) = {P_2}\left( {{x_2},{y_2}} \right)\]
 \[\left( {\dfrac{{ - 2 + 0}}{2},\dfrac{{2 + 5}}{2}} \right)\]
On adding the numerator part we get,
= \[\left( {\dfrac{{ - 2}}{2},\dfrac{7}{2}} \right)\]
= \[\left( { - 1,\dfrac{7}{2}} \right)\]
So, \[{P_1}\left( { - 1,\dfrac{7}{2}} \right)\]
Similarly, to find \[{P_3}\]
Take \[{P_2}{P_3}\] by \[{P_3}B\] is equal to \[\dfrac{k}{k}\]
\[\dfrac{{{P_2}{P_3}}}{{{P_3}B}} = \dfrac{k}{k}\]
Cancelling the terms we get,
\[\dfrac{{{P_2}{P_3}}}{{{P_3}B}} = \dfrac{1}{1}\]
\[{P_2}{P_3}:{P_{}}B = 1:1\]
Hence point \[{P_3}\] divides \[{P_2}B\] into two equal parts
Hence the coordinates of \[{P_3}\] are
\[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)\]
=\[\left( {\dfrac{{0 + 2}}{2},\dfrac{{5 + 8}}{2}} \right)\]
On adding the numerator terms we get,
=\[\left( {\dfrac{2}{2},\dfrac{{13}}{2}} \right)\]
=\[\left( {1,\dfrac{{13}}{2}} \right)\]
So, \[{P_3}\left( {1,\dfrac{{13}}{2}} \right)\]
Hence the coordinates of the points are \[{P_2}(0,5)\], \[{P_1}\left( { - 1,\dfrac{7}{2}} \right)\], \[{P_3}\left( {1,\dfrac{{13}}{2}} \right)\]

Note: Here, without a diagram it is little much difficult to understand the problem. The problem says the line segment which divides the coordinates into four equal parts.
The diagrams show between the coordinates only three points that is \[{P_1},{P_2},{P_3}\] in a blind situation, we mistake four equal parts has \[{P_1},{P_2},{P_3},{P_4}\]. Like this, it is wrong \[A{P_1},{P_1}{P_2},{P_2}{P_3},{P_3}{P_4},{P_4}B\] hence the sum cannot be solved.