Question

The straight line $3x+y=9$ divides the line segment joining the points A$(1,3)$ and B$(2,7)$ in what ratio?

• A. $-3:4$
• B. $3:4$
• C. $4:5$
• D. $-5:6$

Answer 1

The correct option is B $3:4$

Let the given points are

$A(x_1,y_1)=A(1,3)$

$B(x_2,y_2)=B(2,7)$

Let the straight line $3x+y=9$ divides $AB$ in the ratio $K:1$ at point $P(x,y)$

Then using section formula,

$P(x,y)=(\frac{K{x}_2+1x_1}{K+1},\frac{K{y}_1+1y_1}{K+1})$

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

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

As this point $P(x,y)$ lies on the given straight line It will satisfy the lines equation i.e

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

$\Rightarrow \frac{6K+3}{K+1}+\frac{7K+3}{K+1}=9$

$\Rightarrow \frac{6K+3+7K+3}{K+1}=9$

$\Rightarrow 13K+6=9(k+1)$

$\Rightarrow 13K+6=9K+9$

$\Rightarrow 13K-9K=9-6$

$\Rightarrow 4K=3$

$K=\frac{3}{4}$

$\therefore$ the ratio is $3:4$ internally