Find the points of trisection of the line segment joining the points:

Question:

Find the points of trisection of the line segment joining the points:

(a) 5, −6 and (−7, 5),

(b) (3, −2) and (−3, −4),

(c) (2, −2) and (−7, 4).

Solution:

The co-ordinates of a point which divided two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ internally in the ratio $m: n$ is given by the formula,

$(x, y)=\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)$

The points of trisection of a line are the points which divide the line into the ratio.

(i) Here we are asked to find the points of trisection of the line segment joining the points A(5,−6) and B(−7,5).

So we need to find the points which divide the line joining these two points in the ratio and 2 : 1.

Let P(x, y) be the point which divides the line joining ‘AB’ in the ratio 1 : 2.

$(x, y)=\left(\left(\frac{1(-7)+2(5)}{1+2}\right),\left(\frac{1(5)+2(-6)}{1+2}\right)\right)$

$(x, y)=\left(1,-\frac{7}{3}\right)$

Let Q(e, d) be the point which divides the line joining ‘AB’ in the ratio 2 : 1.

$(e, d)=\left(\left(\frac{1(5)+2(-7)}{1+2}\right),\left(\frac{1(-6)+2(5)}{1+2}\right)\right)$

$(e, d)=\left(-3, \frac{4}{3}\right)$

Therefore the points of trisection of the line joining the given points are $\left(1,-\frac{7}{3}\right)$ and $\left(-3, \frac{4}{3}\right)$.

(ii) Here we are asked to find the points of trisection of the line segment joining the points A(3,−2) and B(−3,−4).

So we need to find the points which divide the line joining these two points in the ratio and 2 : 1.

Let P(x, y) be the point which divides the line joining ‘AB’ in the ratio 1 : 2.

$(x, y)=\left(\left(\frac{1(-3)+2(3)}{1+2}\right),\left(\frac{1(-4)+2(-2)}{1+2}\right)\right)$

$(x, y)=\left(1,-\frac{8}{3}\right)$

Let Q(e, d) be the point which divides the line joining ‘AB’ in the ratio 2 : 1

$(e, d)=\left(\left(\frac{1(3)+2(-3)}{1+2}\right),\left(\frac{1(-2)+2(-4)}{1+2}\right)\right)$

$(e, d)=\left(-1,-\frac{10}{3}\right)$

Therefore the points of trisection of the line joining the given points are $\left(1,-\frac{8}{3}\right)$ and $\left(-1,-\frac{10}{3}\right)$.

(iii) Here we are asked to find the points of trisection of the line segment joining the points A(2,−2) and B(−7,4).

So we need to find the points which divide the line joining these two points in the ratio and 2 : 1.

Let P(x, y) be the point which divides the line joining ‘AB’ in the ratio 1 : 2.

$(x, y)=\left(\left(\frac{1(-7)+2(2)}{1+2}\right),\left(\frac{1(4)+2(-2)}{1+2}\right)\right)$

$(x, y)=(-1,0)$

Let Q(e, d) be the point which divides the line joining ‘AB’ in the ratio 2 : 1.

$(e, d)=\left(\left(\frac{1(2)+2(-7)}{1+2}\right),\left(\frac{1(-2)+2(4)}{1+2}\right)\right)$

$(e, d)=(-4,2)$

Therefore the points of trisection of the line joining the given points are $(-1,0)$ and $(-4,2)$.