Question

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Chapter-7 coordinate geometry (exercise 7.1)
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Answer 1

Answer:

1.Let the points (- 1, – 2), (1, 0), ( – 1, 2), and ( – 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

AB= √[(1 + 1)2 + (0 + 2)2]= √[4 + 4] =2√2

BC= √[(-1 – 1)2 + (2 – 0)2]= √[4 + 4]=2√2

CD= √[(-3 + 1)2 + (0 -2)2]= √[4 + 4]=2√2

DA= √[(-3 + 1)2 + (0 -2)2]= √[4 + 4]=2√2

AC= √[(-1 + 1)2 + (2+2)2]= √[0 + 16]=4

BD= √[(-3 – 1)2 + (0 -0)2]= √[0 + 16] =4

Side length = AB = BC = CD = DA = 2√2

Diagonal Measure = AC = BD = 4

Hence, the given points are the vertices of a square

2.Let the points (- 3, 5), (3, 1), (0, 3), and ( – 1, – 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

AB= √[(-3 -3)2 + (1 – 5)2]= √[36 + 16] =2√13

BC= √[(0 – 132 + (3 – 1)2]= √[9 + 4]=√13

CD= √[(-1 – 0)2 + (-4 -3)2]= √[1 + 49]=5√2

DA= √[(-1 + 3)2 + (-4 -5)2]= √[4 + 81]=√85

It is also observed that points A, B and C are collinear.

So, the given points can only form 3 sides i.e, a triangle and not a quadrilateral that has 4 sides.

Hence, the given points cannot form a general quadrilateral.

3.Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

AB= √[(7 -4)2 + (6 – 5)2]= √[9 + 1] =√10

BC= √[(4 – 72 + (3 – 6)2]= √[9 + 9]=√18

CD= √[(1 – 4)2 + (2 -3)2]= √[9 + 1]=√10

DA= √[(1 – 4)2 + (2 -5)2]= √[9 + 9]=√18

AC= √[(4 -4)2 + (3-5)2]= √[0 + 4]=2

BD= √[(1 – 7)2 + (2 -6)2]= √[36 + 16] =2√13

The opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths.

Hence, the given points are the vertices of a parallelogram.