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Q.
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
Straight Lines
Solution:
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{5 K-1}{K+1}, \frac{7 K+1}{K+1}\right)$
Point $P$ will satisfy the line $x+y=4$
i.e., $ \frac{5 K-1}{K+1}+\frac{7 K+1}{K+1} =\frac{4}{1}$
$\Rightarrow \frac{5 K-1+7 K+1}{K+1} =4 $
$\Rightarrow 12 K =4 K+4$
$\Rightarrow 8 K =4 $
$\Rightarrow K =\frac{4}{8}=\frac{1}{2}$
Hence, the required ratio is $K: 1=1: 2$ (internally)
