Question

Point $R(h,k)$ divides a line segment between the axes in the ratio $1:2$. Then, the equation of the line is

• A. $2kx+hy=3hk$
• B. $kx+2hy=3hk$
• C. $2hx+ky=3hk$
• D. $hx+2ky=3hk$

Answer 1

Answer: (A)

Let equation of line $AB$ is
$\frac{x}{a}+\frac{y}{b}=\mathrm{1.....}$(i)

Let a point $R(h,k)$ divide line $AB$ in the ratio $1:2$.
By using section formula, (internal division)
$R(h,k)=\left(\frac{1x_2+2x_1}{1+2},\frac{1y_2+2y_1}{1+2}\right)$
$\Rightarrow h=\frac{1\times 0+2\times a}{1+2}$,
$k=\frac{1\times b+2\times 0}{1+2}$
$[\because P(x,y)$ divide the line $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ in the ratio $m:n$ internally $]$
$\therefore P(x,y)=\left(\frac{m{x}_2+n{x}_1}{n+m},\frac{m{y}_2+n{y}_1}{n+m}\right)]$
$\Rightarrow h=\frac{2a}{3},k=\frac{b}{3}$
$\Rightarrow a=\frac{3h}{2},b=3k$
On putting the values of $a$ and $b$ in Eq. (i), we get
$\frac{x}{\frac{3h}{2}}+\frac{y}{3k}=1$
$\frac{2x}{3h}+\frac{y}{3k}=1$
$\frac{2kx+hy}{3hk}=1$
$2kx+hy=3hk$