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प्रश्न
The points A(–1, –2), B(4, 3), C(2, 5) and D(–3, 0) in that order form a rectangle.
विकल्प
True
False
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उत्तर
This statement is True.
Explanation:
Distance between A(–1, –2), B(4, 3),
AB = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
AB = `sqrt((4 + 1)^2 + (3 + 2)^2`
= `sqrt(5^2 + 5^2)`
= `sqrt(25 + 25)`
= `5sqrt(2)`
Distance between C(2, 5) and D(–3, 0),
CD = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
CD = `sqrt((-3 - 2)^2 + (0 - 5)^2`
= `sqrt((-5)^2 + (-5)^2)`
= `sqrt(25 + 25)`
= `5sqrt(2)`
Distance between A(–1, –2) and D(–3, 0),
AD = `sqrt((-3 + 1)^2 + (0 + 2)^2`
= `sqrt((-2)^2 + 2^2)`
= `sqrt(4 + 4)`
= `2sqrt(2)`
And distance between B(4, 3) and C(2, 5),
BC = `sqrt((4 - 2)^2 + (3 - 5)^2`
= `sqrt(2^2 + (-2)^2)`
= `sqrt(4 + 4)`
= `2sqrt(2)`
We know that, in a rectangle, opposite sides and equal diagonals are equal and bisect each other.
Since, AB = CD and AD = BC
Also, distance between A(–1, –2) and C(2, 5),
AC = `sqrt((2 + 1)^2 + (5 + 2)^2`
= `sqrt(3^2 + 7^2)`
= `sqrt(9 + 49)`
= `sqrt(58)`
And distance between D(–3, 0) and B(4, 3),
DB = `sqrt((4 + 3)^2 + (3 - 0)^2`
= `sqrt(7^2 + 3^2)`
= `sqrt(49 + 9)`
= `sqrt(58)`
Since, diagonals AC and BD are equal.
Hence, the points A(–1, – 2), B(4, 3), C(2, 5) and D(–3, 0) form a rectangle.
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