Question

The straight line $3x+y=9$ divides the line segment joining the points $(1,3)$ and $(2,7)$ in the ratio

• A. $3:4$ externally
• B. $3:4$ internally
• C. $4:5$ internally
• D. $5:6$ externally

Answer 1

The correct option is B

$3:4$ internally

Explanation for the correct option :

By section formula, If a point $(x,y)$ divides the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n,$

Then $(x,y)=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

Let the ratio be $k:1$

Put $(x_1,y_1)=(1,3)$ and $(x_2,y_2)=(2,7)$

We get,

$(x,y)=\left(\frac{2k+1}{k+1},\frac{7k+3}{k+1}\right)$

This point lies on the line $3x+y=9$ . So, It will satisfy the lines equation i.e.

$$\begin{gathered} 3\left(\frac{\left(2k+1\right)}{k+1}\right)+\left(\frac{\left(7k+3\right)}{k+1}\right)=9 \\ \left(\frac{\left(6k+3\right)}{k+1}\right)+\left(\frac{\left(7k+3\right)}{k+1}\right)=9 \\ \left(\frac{6k+3+7k+3}{k+1}\right)=9 \\ 6k+3+7k+3=9k+9 \end{gathered}$$

$$\begin{gathered} \Rightarrow 6k+3+7k+3-9k-9=0 \\ \Rightarrow 4k-3=0 \end{gathered}$$

So $k=\frac{3}{4}$

$$\begin{gathered} \therefore (x,y)=\left(\frac{2\left({\displaystyle \frac{3}{4}}\right)+1}{{\displaystyle \frac{3}{4}}+1},\frac{7\left({\displaystyle \frac{3}{4}}\right)+3}{{\displaystyle \frac{3}{4}}+1}\right) \\ =\left(\frac{{\displaystyle \frac{3}{2}}+1}{{\displaystyle \frac{7}{3}}},\frac{{\displaystyle \frac{21}{4}}+3}{{\displaystyle \frac{7}{3}}}\right) \\ =\left(\frac{{\displaystyle \frac{5}{2}}}{{\displaystyle \frac{7}{3}}},\frac{{\displaystyle \frac{33}{4}}}{{\displaystyle \frac{7}{3}}}\right) \\ =\left(1.07,3.54\right) \end{gathered}$$

As the value of point $(x,y)=(1.07,3.54)$ which divides the line joining the points $(1,3)$ and $(2,7)$ lies in between the coordinates of the line segment then it means it divides the line segment internally.

So the ratio is $3:4$ internally.

Hence the correct option is option B.