Question

If $x-$axis divide the line segment joining the points $(4,6)$ and $(1,-7)$ in ratio $m:n$, then the coordinates of the point of division is

• A. $(\frac{13}{31},0)$
• B. $(\frac{31}{13},0)$
• C. $(\frac{17}{5},0)$
• D. $(\frac{5}{17},0)$

Answer 1

Answer: (B)

$(\frac{31}{13},0)$

Let a point $P(x,y)$ divide the line segment $AB$ joining the points $A(x_1,y_1)\&B(x_2,y_2)$ in the ratio $m:n$.

Here the point is on the $x-$axis.

So, $P(x,y)=(x,0)$

Then, by section formula we have

$x=\frac{n{x}_1+m{x}_2}{m+n}\&y=\frac{n{y}_1+m{y}_2}{m+n}$

Now, $A(x_1,y_1)=(4,6)\&B(x_2,y_2)=(1-7)$

Therefore, $y=\frac{n{y}_1+m{y}_2}{m+n}\Rightarrow 0=\frac{-7m+6n}{m+n}$

$\Rightarrow \frac{m}{n}=\frac{6}{7}$ i.e. $m:n=6:7$

$\Rightarrow x=\frac{n{x}_2+m{x}_1}{m+n}$

$\Rightarrow x=\frac{7\times 1+6\times 4}{6+7}=\frac{31}{13}$

$\Rightarrow P(x,0)=P(\frac{31}{13},0)$

Hence, option B is correct.