Question

Determine the ratio in which the graph of the equation $3x+y=9$ divides the line segment joining the points $A(2,7)$ and $B(1,3)$.

• A. $\frac{4}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{3}{4}$

Answer 1

Answer: (A)

Let $P(x,y)$ be the point which lies on the line representing $3x+y=9$ and dividing $AB$ in the ratio $k:1$

We know that, ia point $P(x,y)$ divides a line segment passing through the points $A(x_1,y_1)$ and $B(x_2,y_2)$ in the ratio $m:n$ then,

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

$\Rightarrow P(x,y)=\frac{k\times 1+1\times 2}{k+1},\frac{k\times 3+1\times 7}{k+1}$

Lets compare the corresponding coordinates,

$\Rightarrow x=\frac{k\times 1+1\times 2}{k+1}$ = $\frac{k+2}{k+1}$

$y=\frac{k\times 3+1\times 7}{k+1}$ = $\frac{3k+7}{k+1}$

Thus point $P$ is $(\frac{k+2}{k+1},\frac{3k+7}{k+1})$

As $P$ lies on $3x+y=9$,

So, $3\left(\frac{k+2}{k+1}\right)+\left(\frac{3k+7}{k+1}\right)=9$

$\Rightarrow 3k+6+3k+7=9k+9$

$\Rightarrow 3k=4$

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

Thus, the required ratio is $4:3$.