Let A(3, 4) and B(k, 7) and midpoint be C(x, y) which lies on the line 2x + 2y +1 = 0
By midpoint formula,
x = \(\frac{x_1 + x_2}2\), y = \(\frac{y_1 + y_2}2\)
For point C(x, y),
x = \(\frac{3 + k}2\), y = \(\frac{4 + 7}2\)....(1)
Here, y = \(\frac{11}2,\)
Hence, substituting value of y in given equation of line,
2x + 2 × \(\frac{11}2 + 1\) = 0
∴ 2x = -12
∴ x = - 6
Now substituting value of x in equation(1), we get.
x = \(\frac{3 + k}2\)
- 6 = \(\frac{3 + k}2\)
∴ -12 = 3 + k
∴ k = -15
Hence, the value of k is -15.