NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.4

Question 1

Step 1: Let the given vertices be \(A(-4,3)\) and \(B(4,3)\). Let the third vertex be \(C(x,y)\).

Step 2: In an equilateral triangle, all sides are equal, so:

\(AC = BC\) and also \(AB = AC\).

Step 3: First find \(AB\).

\(AB^2 = (4-(-4))^2 + (3-3)^2 = 8^2 + 0 = 64\)

So, \(AB = 8\). Hence \(AC = BC = 8\) and therefore \(AC^2 = BC^2 = 64\).

Step 4: Use \(AC = BC\) to get the relation between \(x\) and \(y\).

\(AC^2 = (x+4)^2 + (y-3)^2\) and \(BC^2 = (x-4)^2 + (y-3)^2\).

Since \(AC^2 = BC^2\):

\((x+4)^2 + (y-3)^2 = (x-4)^2 + (y-3)^2\)

Cancel \((y-3)^2\) from both sides:

\((x+4)^2 = (x-4)^2\)

Expand:

\(x^2 + 8x + 16 = x^2 - 8x + 16\)

\(16x = 0 \Rightarrow x = 0\).

Step 5: Now use \(AC = 8\) (i.e., \(AC^2 = 64\)) to find \(y\).

\(AC^2 = (x+4)^2 + (y-3)^2\)

Put \(x=0\):

\(64 = (0+4)^2 + (y-3)^2 = 16 + (y-3)^2\)

\((y-3)^2 = 48\)

\(y-3 = \pm \sqrt{48} = \pm 4\sqrt{3}\)

So \(y = 3 \pm 4\sqrt{3}\).

Step 6: Hence the two possible third vertices are:

\((0,\,3+4\sqrt{3})\) and \((0,\,3-4\sqrt{3})\).

Step 7: Use the given condition: the origin \((0,0)\) lies in the interior of the triangle.

If \(C=(0,3+4\sqrt{3})\), then all vertices have y-coordinate \(\ge 3\), so the triangle lies completely above the x-axis and cannot contain the origin.

Therefore we choose \(C=(0,3-4\sqrt{3})\), which makes the triangle extend downward and include \((0,0)\) inside.

Final Answer: The required third vertex is \(\boxed{(0,\,3-4\sqrt{3})}\).