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Question: Find the coordinates of the points of trisection of the line segment joining the points \(\left( { -...

Find the coordinates of the points of trisection of the line segment joining the points (3,3)\left( { - 3,3} \right) and (3,3)\left( {3, - 3} \right).

Explanation

Solution

Hint: Make a line segment (let’s say AB) with two points names (let’s say P and Q) on it. Mark the points A and B with the coordinates given in the question. Apply the section formula internally i.e. (lx2+mx1l+m,ly2+my1l+m)\left( {\dfrac{{l{x_2} + m{x_1}}}{{l + m}},\dfrac{{l{y_2} + m{y_1}}}{{l + m}}} \right).

Complete step-by-step answer:

Now, when we look at the ratio at which each point divides the line AB internally, we see that it's 1:2 on point P and 2:1 on point Q.
Thus, P divides the line AB internally in the ratio 1:2
Q divides the line AB internally in the ratio 2:1
Writing the section formula internally we have, (lx2+mx1l+m,ly2+my1l+m)\left( {\dfrac{{l{x_2} + m{x_1}}}{{l + m}},\dfrac{{l{y_2} + m{y_1}}}{{l + m}}} \right).
Now, as we have already discussed, P divides the line segment AB in the ratio 1:2 so,
l=1,m=2l = 1,m = 2
And,
(x1,y1)=(3,3)  (x2,y2)=(3,3)  \left( {{x_1},{y_1}} \right) = \left( { - 3,3} \right) \\\ \\\ \left( {{x_2},{y_2}} \right) = \left( {3, - 3} \right) \\\

So, after putting in the values we have till now and the values which were given in the questions we have,
(1×3+2×31+2,1×3+2×31+2)  (363,3+63)  (1,1)  \to \left( {\dfrac{{1 \times 3 + 2 \times - 3}}{{1 + 2}},\dfrac{{1 \times - 3 + 2 \times 3}}{{1 + 2}}} \right) \\\ \\\ \to \left( {\dfrac{{3 - 6}}{3},\dfrac{{ - 3 + 6}}{3}} \right) \\\ \\\ \to \left( { - 1,1} \right) \\\

Since we know that Q divides the line segment AB in the ratio 2:1 so.
l=2,m=1l = 2,m = 1
And,
(x1,y1)=(3,3)  (x2,y2)=(3,3)  \left( {{x_1},{y_1}} \right) = \left( { - 3,3} \right) \\\ \\\ \left( {{x_2},{y_2}} \right) = \left( {3, - 3} \right) \\\

After putting the values again, we have-

(2×3+1×32+1,2×3+1×32+1)  (633,6+33)  (1,1)  \to \left( {\dfrac{{2 \times 3 + 1 \times - 3}}{{2 + 1}},\dfrac{{2 \times - 3 + 1 \times 3}}{{2 + 1}}} \right) \\\ \\\ \to \left( {\dfrac{{6 - 3}}{3},\dfrac{{ - 6 + 3}}{3}} \right) \\\ \\\ \to \left( {1, - 1} \right) \\\

Thus, after evaluating both the points through section formula internally, we now have the coordinates of the points P and Q i.e.-
P(1,1),Q(1,1)\to P\left( { - 1,1} \right),Q\left( {1, - 1} \right)

Note: Whenever asked to find out the coordinates of the point of intersection on any line segment, always apply the section formula and put the values of the coordinates given in the question. Read the coordinates carefully and do not miss the signs of the coordinates for it may vary your answer.