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Mathematics Question on Coordinate Geometry

The vertices of a quadrilateral ABCD are A(6, – 2), B(9, 2), C(5, – 1) and D(2, – 5). Prove that ABCD is a rhombus, and not a square.

Answer

Step 1: Find the lengths of the sides using the distance formula: d=(x2−x1)2+(y2−y1)2.d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. Side AB: AB=(9−6)2+(2−(−2))2=32+42=9+16=5.AB = \sqrt{(9 - 6)^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5. Side BC: BC=(5−9)2+(−1−2)2=(−4)2+(−3)2=16+9=5.BC = \sqrt{(5 - 9)^2 + (-1 - 2)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = 5. Side CD: CD=(2−5)2+(−5−(−1))2=(−3)2+(−4)2=9+16=5.CD = \sqrt{(2 - 5)^2 + (-5 - (-1))^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = 5. Side DA: DA=(6−2)2+(−2−(−5))2=42+32=16+9=5.DA = \sqrt{(6 - 2)^2 + (-2 - (-5))^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5. All sides are equal, so ABCDABCD is a rhombus. Step 2: Check diagonals to confirm it is not a square. Diagonal AC: AC=(5−6)2+(−1−(−2))2=(−1)2+12=1+1=2.AC = \sqrt{(5 - 6)^2 + (-1 - (-2))^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}. Diagonal BD: BD=(2−9)2+(−5−2)2=(−7)2+(−7)2=49+49=98=72.BD = \sqrt{(2 - 9)^2 + (-5 - 2)^2} = \sqrt{(-7)^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2}. Since the diagonals are not equal (AC≠BDAC \neq BD), ABCDABCD is not a square. Correct Answer: ABCDABCD is a rhombus, but not a square.

Explanation

Solution

Step 1: Find the lengths of the sides using the distance formula: d=(x2−x1)2+(y2−y1)2.d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. Side AB: AB=(9−6)2+(2−(−2))2=32+42=9+16=5.AB = \sqrt{(9 - 6)^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5. Side BC: BC=(5−9)2+(−1−2)2=(−4)2+(−3)2=16+9=5.BC = \sqrt{(5 - 9)^2 + (-1 - 2)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = 5. Side CD: CD=(2−5)2+(−5−(−1))2=(−3)2+(−4)2=9+16=5.CD = \sqrt{(2 - 5)^2 + (-5 - (-1))^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = 5. Side DA: DA=(6−2)2+(−2−(−5))2=42+32=16+9=5.DA = \sqrt{(6 - 2)^2 + (-2 - (-5))^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5. All sides are equal, so ABCDABCD is a rhombus. Step 2: Check diagonals to confirm it is not a square. Diagonal AC: AC=(5−6)2+(−1−(−2))2=(−1)2+12=1+1=2.AC = \sqrt{(5 - 6)^2 + (-1 - (-2))^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}. Diagonal BD: BD=(2−9)2+(−5−2)2=(−7)2+(−7)2=49+49=98=72.BD = \sqrt{(2 - 9)^2 + (-5 - 2)^2} = \sqrt{(-7)^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2}. Since the diagonals are not equal (AC≠BDAC \neq BD), ABCDABCD is not a square. Correct Answer: ABCDABCD is a rhombus, but not a square.