Question

In what ratio, the line joining $(-1,1)$ and $(5,7)$ is devided by the line $x+y=4$$?$

Answer 1

Simplification of given data
Let line $AB$ is the line joining the points
$A(-1,1)\&B(5,7)$ and let line $CD$ be $x+y=4$
Let line $AB$ be divided by the line $CD$ at point $P$.
Let $k:1$ be the ratio line $AB$ is divided by the line $CD$.
If a point divides any line joining $(x_1,y_1)$ & $(x_2,y_2)$ in the ratio of $m_1:m_2$ then co-ordinate of that point are $(\frac{m_2{x}_1+m_1{x}_2}{m_1+m_2},\frac{m_2{y}_1+m_1{y}_2}{m_1+m_2})$

Point $P$ which divide the line $A(-1,1)$ & $B(5,7)$ in $k:1$ ratio is
$P=(\frac{\left(5\right)\left(k\right)+(-1)\left(1\right)}{k+1},\frac{\left(7\right)\left(k\right)+\left(1\right)\left(1\right)}{k+1})=(\frac{5k-1}{k+1},\frac{7k+1}{k+1})$

Required point
Now point $P(\frac{5k-1}{k+1},\frac{7k+1}{k+1})$ lies on the line $CD$
So, it will satisfy the equation of line $CD$
So, putting $x=\frac{5k-1}{k+1},y=\frac{7k+1}{k+1}$ in $x+y=4$,

$\Rightarrow \left(\frac{5k-1}{k+1}\right)+\left(\frac{7k+1}{k+1}\right)=4$

$\Rightarrow \frac{(5k-1)+(7k+1)}{k+1}=4$

$\Rightarrow 5k-1+7k+1=4(k+1)$

$\Rightarrow 5k+7k-1+1=4k+4$

$\Rightarrow 12k+0=4k+4$,

$\Rightarrow 12k-4k=4$

$\Rightarrow 8k=4$

$\Rightarrow k=\frac{4}{8}$

$\Rightarrow k=\frac{1}{2}$

Hence, point $P$ divides $AB$ in the ratio of $k:1$

$=\frac{1}{2}:1$

$=2\times \frac{1}{2}:2\times 1$

$=1:2$.