We will discuss here how to use the midpoint formula to find the middle point of a line segment joining the two co-ordinate points.

The coordinates of the midpoint M of a line segment AB with end points A (x\(_{1}\), y\(_{1}\)) and B (y\(_{2}\), y\(_{2}\)) are M (\(\frac{x_{1} + x_{2}}{2}\), \(\frac{y_{1} + y_{2}}{2}\)).

Let M be the midpoint of the line segment joining the points A (x\(_{1}\), y\(_{1}\)) and B (y\(_{2}\), y\(_{2}\)).

Then, M divides AB in the ratio 1 : 1.

So, by the section formula, the coordinates of M are (\(\frac{1\cdot x_{2} + 1\cdot x_{1}}{1 + 1}\)) i.e., (\(\frac{x_{1} + x_{2}}{2}\)).

Therefore, the coordinates of the midpoint of AB are (\(\frac{x_{1} + x_{2}}{2}\), \(\frac{y_{1} + y_{2}}{2}\)).

That is the middle point of the line segment joining the
points (x\(_{1}\), y\(_{1}\)) and (y\(_{2}\), y\(_{2}\)) has the coordinates
(\(\frac{x_{1} + x_{2}}{2}\), \(\frac{y_{1} + y_{2}}{2}\)).

Solved examples on midpoint formula:

**1.** Find the coordinates of the midpoint of the line segment
joining the point A (-5, 4) and B (7, -8).

**Solution:**

Let M (x, y) be the midpoint of AB. Then, x = \(\frac{(-5) + 7}{2}\) = 1 and y = \(\frac{4 + (-8)}{2}\) = -2

Therefore, the required middle point is M (1, -2).

**2.** Let P (6, -3) be the middle point of the line segment AB,
where A has the coordinates (-2, 0). Find the coordinate of B.

**Solution:**

Let the coordinates of B be (m, n). The middle point P on AB has the coordinates (\(\frac{(-2) + m}{2}\), \(\frac{0 + n}{2}\)).

But P has the coordinates (6, -3).

Therefore, \(\frac{(-2) + m}{2}\) = 6 and \(\frac{0 + n}{2}\) = -3

⟹ -2 + m = 12 and n = -6

⟹ m = 12 + 2 and n = -6

Therefore, the coordinates of B (14, -6)

**3.** Find the point A’ if the point A (-3, 4) on reflection in
the point (1, -1) maps onto the point A’.

**Solution:**

Let A’ = (x, y). Clearly, (1, -1) is the middle point of AA’.

The middle point of AA’ = (\(\frac{x + (-3) }{2}\), \(\frac{y + 4}{2}\)) = (1, -1).

⟹ \(\frac{x - 3}{2}\) = 1 and \(\frac{y + 4}{2}\) = -1

⟹ x = 5 and y = -6

Therefore, the coordinate of the point A’ are (5, -6)

● **Distance and Section Formulae**

**Distance Formula****Distance Properties in some Geometrical Figures****Conditions of Collinearity of Three Points****Problems on Distance Formula****Distance of a Point from the Origin****Distance Formula in Geometry****Section Formula****Midpoint Formula****Centroid of a Triangle****Worksheet on Distance Formula****Worksheet on Collinearity of Three Points****Worksheet on Finding the Centroid of a Triangle****Worksheet on Section Formula**

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