Question

If A(1,2),B(4,3) and C(6,6) are the three vertices of a parallelogram ABCD, find the vertex D and also the area.

Answer 1

Answer:

${\pink{\green{\sf{Coordinate\:of\:D=(3,5)}}}}$

${\pink{\green{\sf{Area\:of\:parallelogram=0\:units}}}}$

Step-by-step explanation:

$\huge{\pink{\green{\underline{\red{\sf{SOLUTION-}}}}}}$

》In the given question information given about a parallelogram and its Coordinate of three vertices.

》We have to find the Coordinate of forth vertices and Area of parallelogram.

$\underline \bold{ Given \: coordinates : } \\ \implies A= (1,2) \\ \implies B = (4,3) \\ \implies C = (6,6) \\ \\ \underline \bold{To \: Find : } \\ \implies Coordinate \: of \: D = ( x_{4},y_{4}) \\ \implies Area \: of \: parallelogram = ?$

》According to given question :

$\bold{ For \: mid \: point \: of \: AC \implies } \\ \implies x = \frac{x_{1} +x_{2} }{2} \\ \implies x = \frac{1 + 6}{2} \\ \implies x = \frac{7}{2} \\ \\ \implies y = \frac{y_{1} +y_{2}}{2} \\ \implies y = \frac{2 + 6}{2} \\ \implies y = 4 \\ \\ \bold{For \: mid \: point \: of \: BD \implies} \\ \implies \:x=\frac{x_{3} +x_{4}}{2}\\ \implies \frac{7}{2} = \frac{4 +x_{4}}{2} \\ \implies x_{4} = 7 - 4 \\ \bold{\implies x_{4} = 3} \\ \\ \implies y = \frac{y_{3} +y_{4}}{2} \\ \implies 4 = \frac{3 +y_{4}}{2} \\ \implies y_{4} = 8 - 3 \\ \bold{\implies y_{4} = 5}$

》We know the formula for area of triangle so parallelogram makes two triangles.

$\implies Area \: of \: parallelogram = Area \: of \triangle ABC + Area \: of \triangle ADC \\ \implies Area = \frac{1}{2} (x1 (y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) + \frac{1}{2} (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) )\\ \implies Area = \frac{1}{2} (1(3 - 6) + 4(6 -2) + 6(2 - 3)) + \frac{1}{2} (1(5 - 6) + 3(6 - 2) + 6(2 - 5) )\\ \implies Area = \frac{1}{2} (- 3 + 16 - 6- 1 + 12 - 18) \\ \implies Area = \frac{1}{2} \times 0\\ \bold {\implies area = 0 \:units}$

Answer 2

✿━━━━@♥ℳg━━━━✿

$\boxed{Explained\:Answer}$

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✿━━━━@♥ℳg━━━━✿

$\mathfrak{\huge{\red{ANSWER}}}$

Let A(1, 2), B (4, 3) and C(6, 6).

Let D (x, y) be the fourth vertex of the parallelogram ABCD.

Since, the diagonals of parallelogram bisect each other at O.

∴ Mid point of BD = Mid point of AC

Thus, (3, 3) is the fourth vertex