# In what ratio, the line joining (–1, 1) and (5, 7)

Question:

In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line

x + y = 4?

Solution:

The equation of the line joining the points (–1, 1) and (5, 7) is given by

$y-1=\frac{7-1}{5+1}(x+1)$

$y-1=\frac{6}{6}(x+1)$

$x-y+2=0$ $\ldots(1)$

The equation of the given line is

x + y – 4 = 0 … (2)

The point of intersection of lines (1) and (2) is given by

x = 1 and y = 3

Let point (1, 3) divide the line segment joining (–1, 1) and (5, 7) in the ratio 1:k. Accordingly, by section formula,

$(1,3)=\left(\frac{k(-1)+1(5)}{1+k}, \frac{k(1)+1(7)}{1+k}\right)$

$\Rightarrow(1,3)=\left(\frac{-k+5}{1+k}, \frac{k+7}{1+k}\right)$

$\Rightarrow \frac{-k+5}{1+k}=1, \frac{k+7}{1+k}=3$

$\therefore \frac{-k+5}{1+k}=1$

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

$\Rightarrow 2 k=4$

$\Rightarrow k=2$

Thus, the line joining the points (–1, 1) and (5, 7) is divided by line

x + y = 4 in the ratio 1:2.