NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.2

Question 1

Step 1: Use the distance formula to find side lengths:

For two points \((x_1,y_1)\) and \((x_2,y_2)\),

\(\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).

Step 2: Find the side lengths of \(\triangle ABC\).

• AB: A(-2,0), B(2,0)
  \(AB = \sqrt{(2-(-2))^2 + (0-0)^2} = \sqrt{4^2} = 4\)
• AC: A(-2,0), C(0,2)
  \(AC = \sqrt{(0-(-2))^2 + (2-0)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\)
• BC: B(2,0), C(0,2)
  \(BC = \sqrt{(0-2)^2 + (2-0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\)

Step 3: Find the side lengths of \(\triangle DEF\).

• DE: D(-4,0), E(4,0)
  \(DE = \sqrt{(4-(-4))^2 + (0-0)^2} = \sqrt{8^2} = 8\)
• DF: D(-4,0), F(0,4)
  \(DF = \sqrt{(0-(-4))^2 + (4-0)^2} = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}\)
• EF: E(4,0), F(0,4)
  \(EF = \sqrt{(0-4)^2 + (4-0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{32} = 4\sqrt{2}\)

Step 4: Check ratios of corresponding sides (AB \(\leftrightarrow\) DE, AC \(\leftrightarrow\) DF, BC \(\leftrightarrow\) EF).

• \(\dfrac{DE}{AB} = \dfrac{8}{4} = 2\)
• \(\dfrac{DF}{AC} = \dfrac{4\sqrt{2}}{2\sqrt{2}} = 2\)
• \(\dfrac{EF}{BC} = \dfrac{4\sqrt{2}}{2\sqrt{2}} = 2\)

Step 5: Since all three pairs of corresponding sides are in the same ratio (common ratio = 2), the triangles are similar by the SSS similarity criterion.

Therefore, \(\triangle ABC \sim \triangle DEF\).