If three consecutive vertices of a parallelogram are (1, -2), (3, 6) and (5, 10), find its vertex.

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Solution

let A (1, -2), B (3, 6), C (5, 10) and D(x, y) be the vertices of a parallelogram ABCD taken in order.

Since, the diagonals of a parallelogram bisect each other.

∴ Coordinates of the mid-point of AC = Coordinates of the mid-point of BD
finding mid point of AC
$left parenthesis fraction numerator 1 plus 5 over denominator 2 end fraction comma fraction numerator negative 2 plus 10 over denominator 2 end fraction right parenthesis 6 over 2 comma 8 over 2 left parenthesis 3 comma 2 right parenthesis$

$n o w space 3 equals fraction numerator 3 plus x over denominator 2 end fraction 6 equals 3 plus x x equals 3 s i m i l l a r l y 2 equals fraction numerator 6 plus y over denominator 2 end fraction 4 equals 6 plus y y equals negative 2 h e n c e space t h e space c o o r d i n a r e s space o f space f o u r t h space v e r t e x space i s left parenthesis 3 comma negative 2 right parenthesis space$

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Similar questions

(1, - 2), (3, 6)

(5, 10)

If three consecutive vertices of a parallelogram ABCD areA(1, -2), B(3,6) and C(5,10), find its fourth vertex D.

A B C D

A (1, - 2), B (3, 6)

C (5, 10)

D

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