Question

Find the coordinate of points which trisect the line segment joining (1,-2) and (-3,4).

Answer 1

Hint- Here, we will proceed by firstly finding the ratios in which the points which trisect the line segment joining points (1,-2) and (-3,4) will divide this line segment. Then, we will use the section formula with internal division.

Complete step-by-step answer:

Let AB be a line segment with coordinates of point A as (1,-2) and coordinates of point B as (-3,4) as shown in the figure. Let C and D be the points which trisect the line segment AB.
i.e., AC = CD = DB = $\lambda $ (say)
From the figure, we can write
AD = AC + CD = $\lambda + \lambda = 2\lambda $
Similarly, CB = CD + DB = $\lambda + \lambda = 2\lambda $
$\dfrac{{{\text{AD}}}}{{{\text{DB}}}} = \dfrac{{2\lambda }}{\lambda } = \dfrac{2}{1}$ which means AD:DB = 2:1 i.e., point D divides the line segment AB in the ratio 2:1.
Similarly, $\dfrac{{{\text{AC}}}}{{{\text{CB}}}} = \dfrac{\lambda }{{2\lambda }} = \dfrac{1}{2}$ which means AC:CB = 1:2 i.e., point C divides the line segment AB in the ratio 1:2.
Here, both the points C and D are internal points.
As we know that according to section formula, if a point internally divides a line segment joining the points A(a,b) and B(c,d) in the ratio m:n then the coordinates of that point are given by
x-coordinate of the point = $\dfrac{{\left( {m \times c} \right) + \left( {n \times a} \right)}}{{m + n}}$
y-coordinate of the point = $\dfrac{{\left( {m \times d} \right) + \left( {n \times b} \right)}}{{m + n}}$
For point C, m = 1, n = 2, a = 1, b = -2, c = -3 and d = 4
x-coordinate of point C = $\dfrac{{\left( {1 \times - 3} \right) + \left( {2 \times 1} \right)}}{{1 + 2}} = \dfrac{{ - 3 + 2}}{3} = - \dfrac{1}{3}$
y-coordinate of point C = $\dfrac{{\left( {1 \times 4} \right) + \left( {2 \times - 2} \right)}}{{1 + 2}} = \dfrac{{4 - 4}}{3} = 0$
For point D, m = 2, n = 1, a = 1, b = -2, c = -3 and d = 4
x-coordinate of point D = $\dfrac{{\left( {2 \times - 3} \right) + \left( {1 \times 1} \right)}}{{2 + 1}} = \dfrac{{ - 6 + 1}}{3} = - \dfrac{5}{3}$
y-coordinate of point D = $\dfrac{{\left( {2 \times 4} \right) + \left( {1 \times - 2} \right)}}{{2 + 1}} = \dfrac{{8 - 2}}{3} = \dfrac{6}{3} = 2$
Therefore, the coordinates of the point C is ($ - \dfrac{1}{3}$,0) and the coordinates of the point D is ($ - \dfrac{5}{3}$,2)
Hence, the coordinate of points which trisect the line segment joining (1,-2) and (-3,4) are ($ - \dfrac{1}{3}$,0) and ($ - \dfrac{5}{3}$,2).

Note- In order to trisect a line segment, the points should lie internally between the endpoints of this line segment. We are also having section formula for external division i.e., if a point externally divides a line segment joining the points A(a,b) and B(c,d) in the ratio m:n then the coordinates of that point are given by \[\left[ {\dfrac{{\left( {m \times c} \right) - \left( {n \times a} \right)}}{{m - n}},\dfrac{{\left( {m \times d} \right) - \left( {n \times b} \right)}}{{m - n}}} \right]\].