If the mid-point of the line joining (3, 4) and (k, 7)

We have two points A (3, 4) and B (k, 7) such that its mid-point is.

It is also given that point P lies on a line whose equation is

$2 x+2 y+1=0$

In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as,

$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Therefore mid-point P of side AB can be written as,

$P(x, y)=\left(\frac{k+3}{2}, \frac{7+4}{2}\right)$

Now equate the individual terms to get,

$x=\frac{k+3}{2}$

$y=\frac{11}{2}$

Since, P lies on the given line. So,

$2 x+2 y+1=0$

Put the values of co-ordinates of point P in the equation of line to get,

$2\left(\frac{k+3}{2}\right)+2\left(\frac{11}{2}\right)+1=0$

On further simplification we get,

$k+15=0$

So, $k=-15$