Question

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

Answer 1

Hint: Assume the coordinate of the point C be $\left( {{x}_{1}},{{y}_{1}} \right)$ and the coordinate of the point D be $\left( {{x}_{2}},{{y}_{2}} \right)$ . 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 $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ , $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ . Use this formula and get the coordinates of the points C and D. Then, compare the coordinates of the points C and D with $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ 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 $\left( 0,0 \right)$ and $\left( 9,12 \right)$ .
The coordinate of the point A = $\left( 0,0 \right)$ ………………..……(1)
The coordinate of the point B = $\left( 9,12 \right)$ ……………………(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 $\left( {{x}_{1}},{{y}_{1}} \right)$ and the coordinate of the point D be $\left( {{x}_{2}},{{y}_{2}} \right)$ .
The coordinate of the point C = $\left( {{x}_{1}},{{y}_{1}} \right)$ ………………………..(3)
The coordinate of the point D = $\left( {{x}_{2}},{{y}_{2}} \right)$ ………………………(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 $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ , $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ ………………………(6)
For the line AD, we have, AC = CD. It means that the point C $\left( {{x}_{1}},{{y}_{1}} \right)$ is the midpoint of the line joining the points A $\left( 0,0 \right)$ and D $\left( {{x}_{2}},{{y}_{2}} \right)$ .
Now, using equation (6) to obtain the midpoint of the line AD.
$\left( {{x}_{1}},{{y}_{1}} \right)=\left( \dfrac{0+{{x}_{2}}}{2},\dfrac{0+{{y}_{2}}}{2} \right)$
$\Rightarrow \left( {{x}_{1}},{{y}_{1}} \right)=\left( \dfrac{{{x}_{2}}}{2},\dfrac{{{y}_{2}}}{2} \right)$ ……………………..(7)
Comparing the LHS and RHS of equation (7), we get ${{x}_{1}}=\dfrac{{{x}_{2}}}{2}$ and ${{y}_{1}}=\dfrac{{{y}_{2}}}{2}$ .
Now, solving
${{x}_{1}}=\dfrac{{{x}_{2}}}{2}$
$\Rightarrow 2{{x}_{1}}={{x}_{2}}$ ……………….(8)
Now, solving
${{y}_{1}}=\dfrac{{{y}_{2}}}{2}$
$\Rightarrow 2{{y}_{1}}={{y}_{2}}$ ……………….(9)
For the line CB, we have, CD = DB. It means that the point D $\left( {{x}_{2}},{{y}_{2}} \right)$ is the midpoint of the line joining the points C $\left( {{x}_{1}},{{y}_{1}} \right)$ and B $\left( 9,12 \right)$ .
Now, using equation (6) to obtain the midpoint of the line CB.
$\left( {{x}_{2}},{{y}_{2}} \right)=\left( \dfrac{{{x}_{1}}+9}{2},\dfrac{{{y}_{1}}+12}{2} \right)$ ……………………..(10)
Comparing the LHS and RHS of equation (10), we get ${{x}_{2}}=\dfrac{{{x}_{1}}+9}{2}$ and ${{y}_{2}}=\dfrac{{{y}_{1}}+12}{2}$ .
Now, solving
${{x}_{2}}=\dfrac{{{x}_{1}}+9}{2}$
$\Rightarrow 2{{x}_{2}}={{x}_{1}}+9$ ……………….(11)
Now, putting the value of ${{x}_{2}}$ from equation (8) in equation (10), we get

& \Rightarrow 2\left( 2{{x}_{1}} \right)={{x}_{1}}+9 \\\ & \Rightarrow 4{{x}_{1}}={{x}_{1}}+9 \\\ & \Rightarrow 4{{x}_{1}}-{{x}_{1}}=9 \\\ & \Rightarrow 3{{x}_{1}}=9 \\\ \end{aligned}$$ $$\Rightarrow {{x}_{1}}=3$$ …………………….(12) Putting the value of $${{x}_{1}}$$ in equation (8), we get $$\begin{aligned} & \Rightarrow 2{{x}_{1}}={{x}_{2}} \\\ & \Rightarrow 2.3={{x}_{2}} \\\ \end{aligned}$$ $$\Rightarrow 6={{x}_{2}}$$ ……………………….(13) Now, solving $${{y}_{2}}=\dfrac{{{y}_{1}}+12}{2}$$ $$\Rightarrow 2{{y}_{2}}={{y}_{1}}+12$$ ……………….(14) Now, putting the value of $${{y}_{2}}$$ from equation (9) in equation (14), we get $$\begin{aligned} & \Rightarrow 2\left( 2{{y}_{1}} \right)={{y}_{1}}+12 \\\ & \Rightarrow 4{{y}_{1}}={{y}_{1}}+12 \\\ & \Rightarrow 4{{y}_{1}}-{{y}_{1}}=12 \\\ & \Rightarrow 3{{y}_{1}}=12 \\\ \end{aligned}$$ $$\Rightarrow {{y}_{1}}=4$$ …………………..(15) Putting the value of $${{y}_{1}}$$ in equation (8), we get $$\begin{aligned} & \Rightarrow 2{{y}_{1}}={{y}_{2}} \\\ & \Rightarrow 2.4={{y}_{2}} \\\ \end{aligned}$$ $$\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}}$$ . The coordinate of the point C = $$\left( {{x}_{1}},{{y}_{1}} \right)$$ = $$\left( 3,4 \right)$$ . The coordinate of the point D = $$\left( {{x}_{2}},{{y}_{2}} \right)$$ = $$\left( 6,8 \right)$$ . Hence, the correct option is (C) and (D). Note: We can also solve this question by using section formula. We know the formula that the coordinate of a point which divides the line joining two point $$\left( {{x}_{1}},{{y}_{1}} \right)$$ and $$\left( {{x}_{2}},{{y}_{2}} \right)$$ in the ratio m:n is, $$\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)$$ ……………….(1) Let the coordinate of the point C be $$\left( {{x}_{1}},{{y}_{1}} \right)$$ and the coordinate of the point D be $$\left( {{x}_{2}},{{y}_{2}} \right)$$ . The coordinate of the point C = $$\left( {{x}_{1}},{{y}_{1}} \right)$$ ………………………..(2) The coordinate of the point D = $$\left( {{x}_{2}},{{y}_{2}} \right)$$ ………………………(3)  In the figure we can see that AC = CD = DB. So, we can say that the point C $$\left( {{x}_{1}},{{y}_{1}} \right)$$ is dividing the line AB ratio 1:2 and the point D $$\left( {{x}_{2}},{{y}_{2}} \right)$$is dividing the line AB in the ratio 2:1.  The point C $$\left( {{x}_{1}},{{y}_{1}} \right)$$ is dividing the line joining the points A $$\left( 0,0 \right)$$ and B $$\left( 9,12 \right)$$ in the ratio 1:2. Using, equation (1), we can get the coordinates of the point C $$\left( {{x}_{1}},{{y}_{1}} \right)$$ . $$\begin{aligned} & \left( {{x}_{1}},{{y}_{1}} \right)=\left( \dfrac{1\times 9+2\times 0}{1+2},\dfrac{1\times 12+2\times 0}{1+2} \right) \\\ & \Rightarrow \left( {{x}_{1}},{{y}_{1}} \right)=\left( \dfrac{9}{3},\dfrac{12}{3} \right) \\\ & \Rightarrow \left( {{x}_{1}},{{y}_{1}} \right)=\left( 3,4 \right) \\\ \end{aligned}$$  The point D $$\left( {{x}_{2}},{{y}_{2}} \right)$$ is dividing the line joining the points A $$\left( 0,0 \right)$$ and B $$\left( 9,12 \right)$$ in the ratio 2:1. Using, equation (1), we can get the coordinates of the point D $$\left( {{x}_{2}},{{y}_{2}} \right)$$ . $$\begin{aligned} & \left( {{x}_{2}},{{y}_{2}} \right)=\left( \dfrac{2\times 9+1\times 0}{1+2},\dfrac{2\times 12+1\times 0}{1+2} \right) \\\ & \Rightarrow \left( {{x}_{2}},{{y}_{2}} \right)=\left( \dfrac{18}{3},\dfrac{24}{3} \right) \\\ & \Rightarrow \left( {{x}_{2}},{{y}_{2}} \right)=\left( 6,8 \right) \\\ \end{aligned}$$ The coordinate of the point C = $$\left( {{x}_{1}},{{y}_{1}} \right)$$ = $$\left( 3,4 \right)$$ . The coordinate of the point D = $$\left( {{x}_{2}},{{y}_{2}} \right)$$ = $$\left( 6,8 \right)$$ . Hence, the correct option is (C) and (D).