NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.1

Question 13

Step 1: Recall the midpoint formula.

The midpoint of two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\[ M = \,\left( \dfrac{x_1 + x_2}{2}, \; \dfrac{y_1 + y_2}{2} \right) \]

Step 2: Identify the given points.

Point Q = \((-6, 5)\), Point R = \((-2, 3)\).

Step 3: Find the midpoint of Q and R using the formula.

For the x-coordinate: \(\dfrac{-6 + (-2)}{2} = \dfrac{-8}{2} = -4\).

For the y-coordinate: \(\dfrac{5 + 3}{2} = \dfrac{8}{2} = 4\).

So, midpoint of Q and R = \((-4, 4)\).

Step 4: Compare this midpoint with the given midpoint P.

P is given as \((\dfrac{a}{3}, 4)\).

So, we must have:

\(\dfrac{a}{3} = -4\) and y = 4 (which already matches).

Step 5: Solve for \(a\).

\(\dfrac{a}{3} = -4 \Rightarrow a = -4 \times 3 = -12\).

Final Answer: The value of \(a\) is -12.