Question

The ratio in which the line segment joining $\left(x_1,y_1\right)$ and $(x_2,y_2)$ is divided by $x-$axis is

• A. $y_1:y_2$
• B. $y_2:y_1$
• C. $-y_1:y_2$
• D. $-y_2:y_1$

Answer 1

Given line segment joining the points $(x_1,y_1)$ and $(x_2,y_2)$
Let the point $P(x,y)$ divides the given line segment in the ratio of $k:1$.
Since, we have the section formula that states if the point $P(x,y)$ divides a line segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ in the ration $m:n$ internally then

$P(x,y)=(\frac{n{x}_1+m{x}_2}{m+n},\frac{n{y}_1+m{y}_2}{m+n})$ .........(i)
Since, the ration is $k:1$, substitute $m=k,n=1$ in above equation, we get
$P(x,y)=(\frac{x_1+k{x}_2}{m+n},\frac{y_1+k{y}_2}{m+n})$
Given point $P$ lies on the $x-axis$ , so the $y$-ordinate becomes zero.
Thus, $\frac{y_1+k{y}_2}{m+n}=0$
$\Rightarrow y_1+k{y}_2=0$
$\Rightarrow y_1=-k{y}_2$
$\therefore k=-\frac{y_1}{y_2}$
Therefore, the point $P$ divides the given line segment in the ration of $k:1=-\left(\frac{y_1}{y_2}\right):1$ [By substituting $k$ value]
$=-\frac{y_1}{y_2}$
$=-y_1:y_2$
Hence, the correct option is $C$.