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Coordinate Geometry
CH07 Question Bank
CBSE_CH07_QUESTION_BANK
Grade 10
Question:
Show that the points $A(0,-1)$, $B(2,1)$, $C(0,3)$, $D(-2,1)$ are the vertices of a square. Also find the area of the square.
Step-by-Step Solution
Key Concept: Use the distance formula to show all sides are equal and both diagonals are equal, then use side length squared for the area (not the coordinate area-of-triangle formula).
$AB=\sqrt{(2-0)^2+(1-(-1))^2}=\sqrt{4+4}=2\sqrt2$. [1.0 Mark]
$BC=\sqrt{(0-2)^2+(3-1)^2}=\sqrt{4+4}=2\sqrt2$; $CD=\sqrt{(-2-0)^2+(1-3)^2}=\sqrt{4+4}=2\sqrt2$; $DA=\sqrt{(0-(-2))^2+(-1-1)^2}=\sqrt{4+4}=2\sqrt2$. All four sides are equal. [1.5 Marks]
$AC=\sqrt{(0-0)^2+(3-(-1))^2}=\sqrt{0+16}=4$; $BD=\sqrt{(-2-2)^2+(1-1)^2}=\sqrt{16+0}=4$. Both diagonals are equal. [1.0 Mark]
Since all sides are equal and both diagonals are equal, $ABCD$ is a square. [0.5 Mark]
Area $=(\text{side})^2=(2\sqrt2)^2=8$ square units. [1.0 Mark]
$BC=\sqrt{(0-2)^2+(3-1)^2}=\sqrt{4+4}=2\sqrt2$; $CD=\sqrt{(-2-0)^2+(1-3)^2}=\sqrt{4+4}=2\sqrt2$; $DA=\sqrt{(0-(-2))^2+(-1-1)^2}=\sqrt{4+4}=2\sqrt2$. All four sides are equal. [1.5 Marks]
$AC=\sqrt{(0-0)^2+(3-(-1))^2}=\sqrt{0+16}=4$; $BD=\sqrt{(-2-2)^2+(1-1)^2}=\sqrt{16+0}=4$. Both diagonals are equal. [1.0 Mark]
Since all sides are equal and both diagonals are equal, $ABCD$ is a square. [0.5 Mark]
Area $=(\text{side})^2=(2\sqrt2)^2=8$ square units. [1.0 Mark]
Correct Answer:
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