ABCD is a parallelogram with vertices A (X₁, Y₁) , B(X₂, Y₂) and C (X₃, Y₃). Then the coordinates of the fourth vertex D in terms of the coordinates of A, B and C are

X₁ + X₃ - X₂, Y₁ + Y₃ - Y₂

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X₁ + X₂ - X₃, Y₁ + Y₂ - Y₃

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X₁ - X₂ - X₃, Y₁ - Y₂ - Y₃

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X₁ + X₂ + X₃, Y₁ + Y₂ + Y₃

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Solution

The correct option is A
X₁ + X₃ - X₂, Y₁ + Y₃ - Y₂

Let the coordinates of D be (X, Y). We know that diagonals of the parallelogram dissect each other. Therefore, Mid-point of AC = Mid-point of BD

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Similar questions

Find the centroid of the triangle with vertices A $(x_{1}, y_{1})$ , B $(x_{2}, y_{2})$ and C $(x_{3}, y_{3})$ .

Q. ABCD is a parallelogram with vertices

A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})

. Find the coordinates of the fourth vertex D in terms of

x_{1}, x_{2}, x_{3}, y_{1}, y_{2} and y_{3}

Q. If three points

(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

lie on the same line, prove that

\frac{y_{2} - y_{3}}{x_{2} x_{3}} + \frac{y_{3} - y_{1}}{x_{3} x_{1}} + \frac{y_{1} - y_{2}}{x_{1} x_{2}} = 0

Q. If

A (x_{1}, y_{1}), B x_{2}, y_{2}), C (x_{3}, y_{3})

are the vertices of an equilateral triangle and

(x_{1} - 2)^{2} + (y_{1} - 3)^{2} = (x_{2} - 2)^{2} + (y_{2} - 3)^{2} = (x_{3} - 2)^{2} + (y_{3} - 3)^{2}

, then

The vertices of a square ABCD are given as A (x $_{1}$ , y $_{1}$ ), B (x $_{2}$ , y $_{2}$ ), C (x $_{3}$ , y $_{3}$ ) and D (x $_{4}$ , y $_{4}$ ).
Which of the following will always hold true?

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ABCD is a parallelogram with vertices A (X1, Y1) , B(X2, Y2) and C (X3, Y3). Then the coordinates of the fourth vertex D in terms of the coordinates of A, B and C are

The correct option is A X1 + X3 - X2, Y1 + Y3 - Y2 Let the coordinates of D be (X, Y). We know that diagonals of the parallelogram dissect each other. Therefore, Mid-point of AC = Mid-point of BD

ABCD is a parallelogram with vertices A (X₁, Y₁) , B(X₂, Y₂) and C (X₃, Y₃). Then the coordinates of the fourth vertex D in terms of the coordinates of A, B and C are

The correct option is A
X₁ + X₃ - X₂, Y₁ + Y₃ - Y₂

Let the coordinates of D be (X, Y). We know that diagonals of the parallelogram dissect each other. Therefore, Mid-point of AC = Mid-point of BD