Question

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

Answer 1

Hint: Consider the equation $\dfrac{x}{a}+\dfrac{y}{b}=1$ and use a sectional formula and find the equation of line. The sectional formula for determining the coordinate for a point $M$ is given as:
$M(x,y)=\left[ \left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n} \right),\left( \dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right) \right]$

Complete step-by-step answer:
For a given line with equation $\dfrac{x}{a}+\dfrac{y}{b}=1$
So the figure is as follows,

Let us consider equation of line as $\dfrac{x}{a}+\dfrac{y}{b}=1$ …….. (1)
The intercept form of line is used, it is as follows,
If $a$ and $b$ are non-zero $X$ and $Y$ intercepts of a line $l$ then its equation is of the form
$\dfrac{x}{a}+\dfrac{y}{b}=1$
Since $a$is the X-intercept of the line $l$ , and as we know that if any point lies on the X-axis its value of Y is equal to zero, it passes through the point $A(a,0)$ . Also if $b$ is the Y-intercept of the line $l$ , and we know that any point that lies on the Y-axis has a value of X equal to zero, it passes through the point $B(0,b)$ .
Here $OA=a$ , $OB=b$ and Point $R(h,k)$ divides the line $AB$ in the ratio $1:2$ .
So the Sectional formula is used to determine the coordinate of a point that divides a line into two parts such that the ratio of their length is $m:n$ .
Let $P$ and $Q$ be the given two points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ respectively, and M be the point dividing the line-segment $PQ$ internally in the ratio $m:n$ .
Then form the sectional formula for determining the coordinate for a point $M$ is given as:
$M(x,y)=\left[ \left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n} \right),\left( \dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right) \right]$
Now we know the sectional formula. So applying sectional formula we get,
$P(x,y)=\left[ \left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n} \right),\left( \dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right) \right]$ …….. (2)
Here we get that $P(x,y)=R(h,k)$
Point $A$ $=({{x}_{1}},{{y}_{1}})=(a,0)$
Point $B$ $=({{x}_{2}},{{y}_{2}})=(0,b)$
and $\dfrac{m}{n}=\dfrac{1}{2}$
So Substituting above in (2), We get,
$\begin{align}
 & R(h,k)=\left[ \left( \dfrac{1\times 0+2\times a}{1+2} \right),\left( \dfrac{1\times b+2\times 0}{1+2} \right) \right] \\
 & R(h,k)=\left[ \left( \dfrac{2\times a}{1+2} \right),\left( \dfrac{1\times b}{1+2} \right) \right] \\
\end{align}$
$R(h,k)=\left[ \left( \dfrac{2a}{3} \right),\left( \dfrac{b}{3} \right) \right]$
So here in above equation we get that
$h=\dfrac{2a}{3}$
So simplifying we get $a=\dfrac{3h}{2}$ ……. (3)
$k=\dfrac{b}{3}$
So simplifying we get $b=3k$ …….. (4)
So we have got the values of$a$ and $b$ so substituting (3) and (4) in (1), We get
$\dfrac{x}{\left( \dfrac{3h}{2} \right)}+\dfrac{y}{\left( 3k \right)}=1$
$\dfrac{2x}{3h}+\dfrac{y}{3k}=1$
$\dfrac{6kx+3hy}{9hk}=1$
So simplifying we get,
$\dfrac{2kx+hy}{3hk}=1$
$2kx+hy=3hk$
So $2kx+hy=3hk$ is the required equation of line.

Note: Consideration of the equation is important. First draw a figure so you can easily figure out how to solve it. Sectional Formula should be known $P(x,y)=\left[ \left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n} \right),\left( \dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right) \right]$ . While substituting $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ mostly confusion occurs so avoid it. So while substituting m and n values, confusion should be taken care of and substitution should be done in the order given.