If A (1, 3), B(-1, 2), C(2, 5) and D(x, 4) are the vertices of a ||gm ABCD then the value of x is

(a) 3 (b) 4 (c) 0 (d) $\frac{3}{2}$

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Solution

Let the diagonals AC and BD of the parallelogram bisect each other at point O(X,Y).

Finding point O

$t a k i n g space d i a g o n a l s space A C X equals fraction numerator 1 plus 2 over denominator 2 end fraction equals 3 over 2 Y equals fraction numerator 3 plus 5 over denominator 2 end fraction equals 4 t a k i n g space x space c o space minus o r d i n a t e s space c o n s i d e r i n g space B D X equals 3 over 2 equals fraction numerator x minus 1 over denominator 2 end fraction 3 equals x minus 1 x equals 4$

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If A (1, 3), B(-1, 2), C(2, 5) and D(x, 4) are the vertices of a ||gm ABCD then the value of x is (a) 3 (b) 4 (c) 0 (d) 32

Let the diagonals AC and BD of the parallelogram bisect each other at point O(X,Y). Finding point O

If A (1, 3), B(-1, 2), C(2, 5) and D(x, 4) are the vertices of a ||gm ABCD then the value of x is

(a) 3 (b) 4 (c) 0 (d) $\frac{3}{2}$