Question

The points which trisect the line segment joining the points $$(0,0)$$ and $$(9,12)$$ are
(A) $$(3,4)$$
(B) $$(8,6)$$
(C) $$(6,8)$$
(D) $$(4,0)$$

Answer 1

Hint: Assume the coordinate of the point C be $$(x_1,y_1)$$ and the coordinate of the point D be $$(x_2,y_2)$$ . The point C is the midpoint of the line joining the points A and D, and the point D is the midpoint of the line joining the points C and B. We know the formula for the midpoint of the line joining two points having coordinates $$(x_1,y_1)$$ and $$(x_2,y_2)$$ , $$({\displaystyle \frac{x_1+x_2}{2}},{\displaystyle \frac{y_1+y_2}{2}})$$ . Use this formula and get the coordinates of the points C and D. Then, compare the coordinates of the points C and D with $$(x_1,y_1)$$ and $$(x_2,y_2)$$ respectively. Now, solve it further and get the values of $$x_1$$ , $$y_1$$ , $$x_2$$ , and $$y_2$$ .

Complete step-by-step answer:
According to the question, we have the coordinates of two points which are $$(0,0)$$ and $$(9,12)$$ .
The coordinate of the point A = $$(0,0)$$ ………………..……(1)
The coordinate of the point B = $$(9,12)$$ ……………………(2)
For the line AB to be divided into 3 equal parts we need two more points.
Let the coordinate of the point C be $$(x_1,y_1)$$ and the coordinate of the point D be $$(x_2,y_2)$$ .
The coordinate of the point C = $$(x_1,y_1)$$ ………………………..(3)
The coordinate of the point D = $$(x_2,y_2)$$ ………………………(4)

The points C and D are trisecting the line AB. We can say that AC is equal to CD and CD is equal to DB. So,
AC = CD = DB …………………………(5)
We know the formula for the midpoint of the line joining two points having coordinates $$(x_1,y_1)$$ and $$(x_2,y_2)$$ , $$({\displaystyle \frac{x_1+x_2}{2}},{\displaystyle \frac{y_1+y_2}{2}})$$ ………………………(6)
For the line AD, we have, AC = CD. It means that the point C $$(x_1,y_1)$$ is the midpoint of the line joining the points A $$(0,0)$$ and D $$(x_2,y_2)$$ .
Now, using equation (6) to obtain the midpoint of the line AD.
$$(x_1,y_1)=({\displaystyle \frac{0+x_2}{2}},{\displaystyle \frac{0+y_2}{2}})$$
$$\Rightarrow (x_1,y_1)=({\displaystyle \frac{x_2}{2}},{\displaystyle \frac{y_2}{2}})$$ ……………………..(7)
Comparing the LHS and RHS of equation (7), we get $$x_1={\displaystyle \frac{x_2}{2}}$$ and $$y_1={\displaystyle \frac{y_2}{2}}$$ .
Now, solving
$$x_1={\displaystyle \frac{x_2}{2}}$$
$$\Rightarrow 2x_1=x_2$$ ……………….(8)
Now, solving
$$y_1={\displaystyle \frac{y_2}{2}}$$
$$\Rightarrow 2y_1=y_2$$ ……………….(9)
For the line CB, we have, CD = DB. It means that the point D $$(x_2,y_2)$$ is the midpoint of the line joining the points C $$(x_1,y_1)$$ and B $$(9,12)$$ .
Now, using equation (6) to obtain the midpoint of the line CB.
$$(x_2,y_2)=({\displaystyle \frac{x_1+9}{2}},{\displaystyle \frac{y_1+12}{2}})$$ ……………………..(10)
Comparing the LHS and RHS of equation (10), we get $$x_2={\displaystyle \frac{x_1+9}{2}}$$ and $$y_2={\displaystyle \frac{y_1+12}{2}}$$ .
Now, solving
$$x_2={\displaystyle \frac{x_1+9}{2}}$$
$$\Rightarrow 2x_2=x_1+9$$ ……………….(11)
Now, putting the value of $$x_2$$ from equation (8) in equation (10), we get
$$\begin{array}{rl}& \Rightarrow 2\left(2x_1\right)=x_1+9\\ & \Rightarrow 4x_1=x_1+9\\ & \Rightarrow 4x_1-x_1=9\\ & \Rightarrow 3x_1=9\end{array}$$
$$\Rightarrow x_1=3$$ …………………….(12)
Putting the value of $$x_1$$ in equation (8), we get
$$\begin{array}{rl}& \Rightarrow 2x_1=x_2\\ & \Rightarrow 2.3=x_2\end{array}$$
$$\Rightarrow 6=x_2$$ ……………………….(13)
 Now, solving
$$y_2={\displaystyle \frac{y_1+12}{2}}$$
$$\Rightarrow 2y_2=y_1+12$$ ……………….(14)
Now, putting the value of $$y_2$$ from equation (9) in equation (14), we get
$$\begin{array}{rl}& \Rightarrow 2\left(2y_1\right)=y_1+12\\ & \Rightarrow 4y_1=y_1+12\\ & \Rightarrow 4y_1-y_1=12\\ & \Rightarrow 3y_1=12\end{array}$$
$$\Rightarrow y_1=4$$ …………………..(15)
Putting the value of $$y_1$$ in equation (8), we get
$$\begin{array}{rl}& \Rightarrow 2y_1=y_2\\ & \Rightarrow 2.4=y_2\end{array}$$
$$\Rightarrow 8=y_2$$ ………………..(16)
From equation (12), equation (13), equation (15), and equation (16), we have the values of $$x_1$$ , $$y_1$$ , $$x_2$$ , and $$y_2$$ .
The coordinate of the point C = $$(x_1,y_1)$$ = $$(3,4)$$ .
The coordinate of the point D = $$(x_2,y_2)$$ = $$(6,8)$$ .
Hence, the correct option is (C) and (D).

Note: We can also solve this question by using section formula.
We know the formula that the coordinate of a point which divides the line joining two point
$$(x_1,y_1)$$ and $$(x_2,y_2)$$ in the ratio m:n is, $$({\displaystyle \frac{m{x}_2+n{x}_1}{m+n}},{\displaystyle \frac{m{y}_2+n{y}_1}{m+n}})$$ ……………….(1)
Let the coordinate of the point C be $$(x_1,y_1)$$ and the coordinate of the point D be $$(x_2,y_2)$$ .
The coordinate of the point C = $$(x_1,y_1)$$ ………………………..(2)
The coordinate of the point D = $$(x_2,y_2)$$ ………………………(3)

In the figure we can see that AC = CD = DB. So, we can say that the point C $$(x_1,y_1)$$ is dividing the line AB ratio 1:2 and the point D $$(x_2,y_2)$$is dividing the line AB in the ratio 2:1.

The point C $$(x_1,y_1)$$ is dividing the line joining the points A $$(0,0)$$ and B $$(9,12)$$ in the ratio 1:2.
Using, equation (1), we can get the coordinates of the point C $$(x_1,y_1)$$ .
$$\begin{array}{rl}& (x_1,y_1)=({\displaystyle \frac{1\times 9+2\times 0}{1+2}},{\displaystyle \frac{1\times 12+2\times 0}{1+2}})\\ & \Rightarrow (x_1,y_1)=({\displaystyle \frac{9}{3}},{\displaystyle \frac{12}{3}})\\ & \Rightarrow (x_1,y_1)=(3,4)\end{array}$$

The point D $$(x_2,y_2)$$ is dividing the line joining the points A $$(0,0)$$ and B $$(9,12)$$ in the ratio 2:1.
Using, equation (1), we can get the coordinates of the point D $$(x_2,y_2)$$ .
$$\begin{array}{rl}& (x_2,y_2)=({\displaystyle \frac{2\times 9+1\times 0}{1+2}},{\displaystyle \frac{2\times 12+1\times 0}{1+2}})\\ & \Rightarrow (x_2,y_2)=({\displaystyle \frac{18}{3}},{\displaystyle \frac{24}{3}})\\ & \Rightarrow (x_2,y_2)=(6,8)\end{array}$$
The coordinate of the point C = $$(x_1,y_1)$$ = $$(3,4)$$ .
The coordinate of the point D = $$(x_2,y_2)$$ = $$(6,8)$$ .
Hence, the correct option is (C) and (D).