Question

Find the coordinates of the points of trisection of the line segment joining (4,-1)and(-2,-3)
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Answer 1

Well, what should be meant by trisection of a line?

Trisection simply means dividing a line segment into three equal parts internally. So a line segment has two points of trisection for getting it into three equal parts, like:

• a line segment has one point of bisection to get it into two equal parts.
• a line segment has two points of trisection to get it into three equal parts, as discussed above.
• a line segment has three points of division to get it into four equal parts.

In general we say that a line segment contains (n - 1) points for getting it into 'n' equal parts.

Okay, but what about the ratio by which the line segment is divided into 'n' equal parts?

Suppose a line segment of length 'n' units is divided into 'n' equal parts of 1 unit each, so the line contains (n - 1) points of division to get it into 'n' equal parts. Let me take the p'th point among the (n - 1) ones arbitrarily, counted from one end of the line segment.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(0,0)(35,0){3}{\line(1,0){15}}\multiput(0,0)(35,0){3}{\multiput(5,0)(5,0){2}{\circle*{1}}}\multiput(0,0)(35,0){2}{\multiput(20,0)(5,0){3}{\circle*{0.5}}}\put(40,5){\vector(-1,0){40}}\put(45,5){\vector(1,0){40}}\put(41.5,4.5){ n }\put(4,-4.5){ 1 }\put(9,-4.5){ 2 }\put(44,-4.5){ p }\put(76,-4.5){ n-1 }\put(20,-7.5){\vector(-1,0){20}}\put(25,-7.5){\vector(1,0){20}}\put(21.5,-8){ p }\end{picture}$

Well, the distance between the first point of division and the end of the line segment, from where the point is counted and taken as first point of division, will be 1 unit, whereas the distance between the point and the other end of the segment will be (n - 1) units.

Similarly, the distance between the p'th point of division and the end of the line segment, from where the point is counted and taken as p'th point, will be 'p' units, whereas the distance between the point and the other end of the segment will be (n - p) units.

Thus we can say that the p'th point divides the line segment in the ratio p : n - p, measured from the end of the segment where the point of division is got to be p'th.

In our case n = 3 and p can be 1 or 2, since there are 2 points of trisection.

On applying the same logic, the 1st point divides the segment in the ratio 1 : 2 and the 2nd point in the ratio 2 : 1.

Case 1:

$\setlength{\unitlength}{1cm}\begin{picture}(5,5)\multiput(4,-1)(-6,-2){2}{\circle*{0.1}}\put(-2,-3){\line (3,1){6}}\put(4.3,-1){\small(4,-1)}\put(-3.5,-3){\small(-2,-3)}\put(2,-1.7){\circle*{0.1}}\put(1,-1.5){\small(x,\ y)}\put(0.2,-2.7){ 2 }\put(3.3,-1.6){\small1}\end{picture}$

Here (x, y) is the point which divides the line segment in the ratio 2 : 1 measured from the point (-2, -3). By section formula,

$x=\dfrac {4\times2+(-2)\times1}{1+2}\\\\\\x=\dfrac {8-2}{3}\\\\\\x=2$

And,

$y=\dfrac {(-1)\times2+(-3)\times1}{1+2}\\\\\\y=\dfrac {-2-3}{3}\\\\\\y=-\dfrac {5}{3}$

So the point is,

$(x,\ y)=\left (2,\ -\dfrac {5}{3}\right)$

Case 2:

$\setlength{\unitlength}{1cm}\begin{picture}(5,5)\multiput(4,-1)(-6,-2){2}{\circle*{0.1}}\put(-2,-3){\line (3,1){6}}\put(4.3,-1){\small(4,-1)}\put(-3.5,-3){\small(-2,-3)}\put(0,-2.33){\circle*{0.1}}\put(-1.2,-2.3){\small(x,\ y)}\put(2,-2){ 2 }\put(-1,-3){\small1}\end{picture}$

Here (x, y) is the point which divides the line segment in the ratio 1 : 2 measured from the point (-2, -3). By section formula,

$x=\dfrac {4\times1+(-2)\times2}{1+2}\\\\\\x=\dfrac {4-4}{3}\\\\\\x=0$

And,

$y=\dfrac {(-1)\times1+(-3)\times2}{1+2}\\\\\\y=\dfrac {-1-6}{3}\\\\\\y=-\dfrac {7}{3}$

So the point is,

$(x,\ y)=\left (0,\ -\dfrac {7}{3}\right)$

Finally, the points of trisection are,

$\Large\boxed{\mathbf{\left (2,\ -\dfrac {5}{3}\right)\ and\ \left (0,\ -\dfrac {7}{3}\right)}}$

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Answer 2

Answer:

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