Prove that the coordinates, x and y, of the middle point of the line joining the point (2,3) to the point (3, 4) satisfy the equation
$x - y + 1 = 0$ .

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Solution

Let $(2, 3)$ and $(3, 4)$ Let us have point $R (x, y)$ be the mid-point of line joining $P$ and $Q$
$\Rightarrow$ Ratio $= 1 : 1$
Applying section formula
$x = \frac{2 \times 1 + 3 \times 1}{1 + 1}$
$\Rightarrow x = \frac{5}{2}$ and
$y = \frac{3 \times 1 + 4 \times 1}{1 + 1}$
$\Rightarrow y = \frac{7}{2}$
To check, $R (\frac{5}{2}, \frac{7}{2})$ satisfies the given equation i.e $x - y + 1 = 0$

$\Rightarrow L H S = \frac{5}{2} - \frac{7}{2} + 1$
$= - 1 + 1 = 0$

$\Rightarrow L H S = R H S = 0$
Hence, $R (\frac{5}{2}, \frac{7}{2})$ satisfy the given equation.

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Prove that the coordinates, x and y, of the middle point of the line joining the point (2,3) to the point (3, 4) satisfy the equation x−y+1=0.

Prove that the coordinates, x and y, of the middle point of the line joining the point (2,3) to the point (3, 4) satisfy the equation
$x - y + 1 = 0$ .