CBSE Class 10 Mathematics Basic 2023: question 31

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Mid-point of AC $=\left(\frac{-1+2}{2},\frac{0+2}{2}\right)=\left(\frac{1}{2},1\right)$

   ½ mark

2. Mid-point of BD $=\left(\frac{3-2}{2},\frac{1+1}{2}\right)=\left(\frac{1}{2},1\right)$

   ½ mark

3. The diagonals AC and BD have the same mid-point, i.e. they bisect each other, so ABCD is a parallelogram.

   1 mark

4. $AC=\sqrt{(2+1{)}^2+(2-0{)}^2}=\sqrt{13}$, $BD=\sqrt{(-2-3{)}^2+(1-1{)}^2}=\sqrt{25}=5$. Since $AC\ne BD$, the diagonals are not equal, so ABCD is not a rectangle.

   1 mark

Answer: ABCD is a parallelogram (diagonals bisect each other) but not a rectangle ($AC=\sqrt{13}\ne BD=5$).

Revise the chapter

• Chapter: Coordinate Geometry (notes, formulas, where marks are lost and every past question)
• Topic: Distance Formula