Question

Q.4 show that A(4, -1), B(6, 0), C(7, -2), and D(5, -3) are the vertices of a square. Please explain it

Answer 1

Answer:

Distance Formula

$= \sqrt{ {(x2 - x1)}^{2} } + (y2 - y1)}^{2}$

(a) Distance between A and B

$= \sqrt{ {(6 - 4)}^{2} } + {(0 + 1)}^{2}$

$= \sqrt{ {2}^{2} } + {1}^{2}$

$\sqrt{4 + 1}$

$= \sqrt{5}$

(b) Distance between B and C

$= \sqrt{ {(7 - 6)}^{2} } + {( - 2 - 0)}^{2}$

$= \sqrt{ {(1)}^{2} } + {( - 2)}^{2}$

$= \sqrt{1 + 4}$

$= \sqrt{5}$

(c) Distance between C and D

$= \sqrt{ {(5 - 7)}^{2} } + {( - 3 + 2)}^{2}$

$= \sqrt{ { ( - 2)}^{2} } + {( - 1)}^{2}$

$= \sqrt{4 + 1}$

$= \sqrt{5}$

(d) Distance between A and D

$= \sqrt{ {(5 - 4)}^{2} } + {( - 3 + 1)}^{2}$

$= \sqrt{ {(1)}^{2} } + {( - 2)}^{2}$

$= \sqrt{1 + 4}$

$= \sqrt{5}$

As the distance between A and B, B and C, C and D and A and D is equal

Therefore, ABCD is a square.

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