Question

The line joining $(-1,1)$ and $(5,7)$ is divided by the line $x+y=4$ in the ratio $K:1$, where $K$ stands for

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

Answer 1

Answer: (D)

Let the point $P$ divides the line joining the points $A(-1,1)$ and $B(5,7)$ in the ratio is $K:1$.

Then,
$P=\left(\frac{K\times x_2+1\times x_1}{K+1},\frac{K\times y_2+1\times y_1}{K+1}\right)$
$(\because x_1=-1,y_1=1,x_2=5,y_2=7)$
$=\left(\frac{K\times 5+1\times (-1)}{K+1},\frac{K\times 7+1\times 1}{K+1}\right)$
$=\left(\frac{5K-1}{K+1},\frac{7K+1}{K+1}\right)$
Point $P$ will satisfy the line $x+y=4$
i.e., $\frac{5K-1}{K+1}+\frac{7K+1}{K+1}=\frac{4}{1}$
$\Rightarrow \frac{5K-1+7K+1}{K+1}=4$
$\Rightarrow 12K=4K+4$
$\Rightarrow 8K=4$
$\Rightarrow K=\frac{4}{8}=\frac{1}{2}$
Hence, the required ratio is $K:1=1:2$ (internally)