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Home » If the three consecutive vertices of a parallelogram be A(3, 4, -3), B(7, 10, -3) and C(5, -2, 7), find the fourth vertex D.
If the three consecutive vertices of a parallelogram be A(3, 4, -3), B(7, 10, -3) and C(5, -2, 7), find the fourth vertex D.
CBSE | Class 11 | Exercise 26C | Maths | RS Aggarwal | Three Dimensional Geometry
If the three consecutive vertices of a parallelogram be A(3, 4, -3), B(7, 10, -3) and C(5, -2, 7), find the fourth vertex D.

Answer:

The vertices of the parallelogram be A(3, 4, -3), B(7, 10, -3) and C(5, -2, 7), and the fourth coordinate be D(a,b,c).

The property of parallelogram is the diagonal bisect each other.

The diagonal AC and BD will bisect each other, and the bisecting point will be equal to the two diagnals.

Formula –

Using the above formula,

The coordinate of mid point of diagonal AC,

\begin{array}{l} = \left( {\frac{{1 \times 5 + 1 \times 3}}{{1 + 1}},\frac{{1 \times -2 + 1 \times 4}}{{1 + 1}},\frac{{1 \times 7 + 1 \times -3}}{{1 + 1}}} \right)\\ = (4, 1 ,2) \end{array}

The coordinate of mid point of diagonal BD,

\begin{array}{l} = \left( {\frac{{1 \times a + 1 \times 7}}{{1 + 1}},\frac{{1 \times b + 1 \times 10}}{{1 + 1}},\frac{{1 \times c + 1 \times -3}}{{1 + 1}}} \right)\\ = \left( {\frac{{a + 7}}{2},\frac{{b + 10}}{2},\frac{c - 3}{2}} \right) \end{array}

Equating the two mid points,

\begin{array}{l} (4, 1, 2) = \left( {\frac{{a + 7}}{2},\frac{{b + 10}}{2},\frac{c - 3}{2}} \right) \end{array}

\begin{array}{l} 4 = \left {\frac{{a + 7}}{2} \end{array}

\begin{array}{l} 1 = \left {\frac{{b + 10}}{2} \end{array}

\begin{array}{l} 2 = \left {\frac{c - 3}{2} \end{array}

8 = a + 7

a = 1

b + 10 = 2

b = -8

c = 7

The point is ( 1, -8, 7).

 

 

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