Show that,(1, -3/2),(-3, -7/2) and (-4, -3/2) | vedclass

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(A) Let the vertices be $A(1, -3/2)$,$B(-3, -7/2)$,and $C(-4, -3/2)$.To check if it is a right-angled triangle,we calculate the squares of the lengths

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(A) Let the vertices be $A(1, -3/2)$,$B(-3, -7/2)$,and $C(-4, -3/2)$.
To check if it is a right-angled triangle,we calculate the squares of the lengths of the sides using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.
$AB^2 = (-3 - 1)^2 + (-7/2 - (-3/2))^2 = (-4)^2 + (-4/2)^2 = 16 + (-2)^2 = 16 + 4 = 20$.
$BC^2 = (-4 - (-3))^2 + (-3/2 - (-7/2))^2 = (-1)^2 + (4/2)^2 = 1 + 2^2 = 1 + 4 = 5$.
$AC^2 = (-4 - 1)^2 + (-3/2 - (-3/2))^2 = (-5)^2 + (0)^2 = 25 + 0 = 25$.
Since $AB^2 + BC^2 = 20 + 5 = 25$,which is equal to $AC^2$,the triangle satisfies the Pythagorean theorem.
Therefore,the given points form a right-angled triangle.

Show that,$(1, -3/2)$,$(-3, -7/2)$ and $(-4, -3/2)$ are the vertices of a right-angled triangle.

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