All these RS Aggarwal class 10 solutions Chapter 16 Exercise: 16.2 Coordinate Geometry are solved by Byju's top ranked professors as per CBSE guidelines.

**Q1) Show that the following points are the vertices of a square: **

**(i) A (3, 2),B(0, 5), C(-3, 2) and D(0, -1) **

**Solution**

**i)**The given points are A (3,2), B(0,5) and C(-3,2), D(0,-1). Then

AB = \(\sqrt{\left ( 0-3 \right )^{2}+\left ( 5-2 \right )^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( 3 \right )^{2}}\)

= \(\sqrt{9+9}\)

= \(\sqrt{18}\)

= \(3\sqrt{2}\) units

BC = \(\sqrt{\left (-3-0 \right )^{2}+\left ( 2-5 \right )^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( -3 \right )^{2}}\)

= \(\sqrt{9+9}\)

= \(\sqrt{18}\)

= \(3\sqrt{2}\) units

CD = \(\sqrt{\left ( 0+3 \right )^{2}+\left ( -1-2 \right )^{2}}\)

= \(\sqrt{\left (3\right )^{2}+\left ( -3 \right )^{2}}\)

= \(\sqrt{9+9}\)

= \(\sqrt{18}\)

= \(3\sqrt{2}\) units

DA = \(\sqrt{\left ( 0-3 \right )^{2}+\left ( -1-2 \right )^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( -3 \right )^{2}}\)

= \(\sqrt{9+9}\)

= \(\sqrt{18}\)

= \(3\sqrt{2}\) units

Therefore AB = BC = CD = DA = \(3\sqrt{2}\) units

Also, AC = \(\sqrt{\left ( -3-3 \right )^{2}+\left ( 2-2 \right )^{2}}\)

= \(\sqrt{\left (-6\right )^{2}+\left ( 0 \right )^{2}}\)

= \(\sqrt{36}\)

= 6 units

BD = \(\sqrt{\left ( 0-0 \right )^{2}+\left ( -1-5 \right )^{2}}\)

= \(\sqrt{\left (0\right )^{2}+\left ( -6 \right )^{2}}\)

= \(\sqrt{36}\)

= 6 units

Thus, diagonal AC = diagonal BD

Therefore, the given points form a square.

**ii) A(6, 2), B(2, 1), C(1, 5) and D(5, 6)**

**Solution**

The given points are A (6,2), B(2,1) and C(1,5), D(5,6). Then

AB = \(\sqrt{\left ( 2-6 \right )^{2}+\left ( 1-2 \right )^{2}}\)

= \(\sqrt{\left (-4\right )^{2}+\left ( -1 \right )^{2}}\)

= \(\sqrt{16+1}\)

= \(\sqrt{17}\) units

BC = \(\sqrt{\left (1-2 \right )^{2}+\left ( 5-1 \right )^{2}}\)

= \(\sqrt{\left (-1\right )^{2}+\left ( -4 \right )^{2}}\)

= \(\sqrt{1+16}\)

= \(\sqrt{17}\) units

CD = \(\sqrt{\left ( 5-1 \right )^{2}+\left ( 6-5 \right )^{2}}\)

= \(\sqrt{\left (4\right )^{2}+\left ( 1 \right )^{2}}\)

= \(\sqrt{16+1}\)

= \(\sqrt{17}\) units

DA = \(\sqrt{\left ( 5-6 \right )^{2}+\left ( 6-2 \right )^{2}}\)

= \(\sqrt{\left (1\right )^{2}+\left ( 4 \right )^{2}}\)

= \(\sqrt{1+16}\)

= \(\sqrt{17}\) units

Therefore AB = BC = CD = DA = \(\sqrt{17}\) units

Also, AC = \(\sqrt{\left ( 1-6 \right )^{2}+\left ( 5-2 \right )^{2}}\)

= \(\sqrt{\left (-5\right )^{2}+\left ( 3 \right )^{2}}\)

= \(\sqrt{25+9}\)

=\(\sqrt{34}\) units

BD = \(\sqrt{\left ( 5-2 \right )^{2}+\left ( 6-1 \right )^{2}}\)

= \(\sqrt{\left (3\right )^{2}+\left ( 5 \right )^{2}}\)

= \(\sqrt{9+25}\)

= \(\sqrt{34}\) units

Thus, diagonal AC = diagonal BD

Therefore, the given points form a square.

**iii)A(0, -2), B ( 3, 1), C (0, 4) and D (-3, 1)**

**Solution**

The given points are P (0,-2), Q(3,1) and R(0,4), S(-3,1). Then

PQ = \(\sqrt{\left ( 3-0 \right )^{2}+\left ( 1+2 \right )^{2}}\)

= \(\sqrt{\left (3\right )^{2}+\left ( 3 \right )^{2}}\)

= \(\sqrt{9+9}\)

= \(\sqrt{18}\)

= \(3\sqrt{2}\) units

QR = \(\sqrt{\left (0-3 \right )^{2}+\left ( 4-1 \right )^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( 3 \right )^{2}}\)

= \(\sqrt{9+9}\)

= \(\sqrt{18}\)

= \(3\sqrt{2}\) units

RS = \(\sqrt{\left ( -3-0 \right )^{2}+\left ( 1-4 \right )^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( -3 \right )^{2}}\)

= \(\sqrt{9+9}\)

= \(\sqrt{18}\)

= \(3\sqrt{2}\) units

SP = \(\sqrt{\left ( -3-0 \right )^{2}+\left ( 1+2 \right )^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( 3 \right )^{2}}\)

= \(\sqrt{9+9}\)

= \(\sqrt{18}\)

= \(3\sqrt{2}\) units

Therefore PQ = QR = RS = SP = \(3\sqrt{2}\) units

Also, PR = \(\sqrt{\left ( 0-0\right )^{2}+\left ( 4+2 \right )^{2}}\)

= \(\sqrt{\left (0\right )^{2}+\left ( 6 \right )^{2}}\)

= \(\sqrt{36}\)

= 6 units

QS = \(\sqrt{\left ( -3-3 \right )^{2}+\left ( 1-1 \right )^{2}}\)

= \(\sqrt{\left (-6\right )^{2}+\left ( 0 \right )^{2}}\)

= \(\sqrt{36}\)

= 6 units

Thus, diagonal PR = diagonal QS

Therefore, the given points form a square.

**Q2) Show that the points A(-3,2), B(-5,-5), C(2,-3) and D(4,4) are the vertices of a rhombus. Find the area of this rhombus.**

The given points are A (-3,2), B(-5,-5) and C(2,-3), D(4,4). Then

AB = \(\sqrt{\left ( -5+3 \right )^{2}+\left ( -5-2 \right )^{2}}\)

= \(\sqrt{\left (-2\right )^{2}+\left ( -7 \right )^{2}}\)

= \(\sqrt{4+49}\)

= \(\sqrt{53}\) units

BC = \(\sqrt{\left (2+5 \right )^{2}+\left ( -3+5 \right )^{2}}\)

= \(\sqrt{\left (7\right )^{2}+\left ( 2 \right )^{2}}\)

= \(\sqrt{49+4}\)

= \(\sqrt{53}\) units

CD = \(\sqrt{\left ( 4-2 \right )^{2}+\left ( 4+3 \right )^{2}}\)

= \(\sqrt{\left (2\right )^{2}+\left ( 7 \right )^{2}}\)

= \(\sqrt{4+49}\)

= \(\sqrt{53}\) units

DA = \(\sqrt{\left ( 4+3 \right )^{2}+\left ( 4-2 \right )^{2}}\)

= \(\sqrt{\left (7\right )^{2}+\left ( 2 \right )^{2}}\)

= \(\sqrt{49+4}\)

= \(\sqrt{53}\) units

Therefore AB = BC = CD = DA = \(\sqrt{53}\) units

Also, AC = \(\sqrt{\left ( 2+3 \right )^{2}+\left ( -3-2 \right )^{2}}\)

= \(\sqrt{\left (5\right )^{2}+\left ( -5 \right )^{2}}\)

= \(\sqrt{25+25}\)

= \(\sqrt{50}\)

= \(\sqrt{25*2}\)

= \(5\sqrt{2}\) units

BD = \(\sqrt{\left ( 4+5 \right )^{2}+\left ( 4+5 \right )^{2}}\)

= \(\sqrt{\left (9\right )^{2}+\left ( 9 \right )^{2}}\)

= \(\sqrt{81+81}\)

= \(\sqrt{162}\)

= \(\sqrt{81*2}\)

= \(9\sqrt{2}\) units

Thus, diagonal AC is not equal to diagonal BD

Therefore, ABCD is a quadrilateral with equal sides and unequal diagonals

Hence, ABCD is a rhombus

Area of a rhombus = \(\frac{1}{2}*\left ( Product\;of\;its\;diagonals\right )\)

= \(\frac{1}{2}*5\sqrt{2}*9\sqrt{2}\)

= \(\frac{45*2}{2}\)

= 45 square units

**Q3) Show that the points A(3,0), B(4,5), C(-1,4) and D(-2,-1) are the vertices of a rhombus. Find its area.**

The given points are A (3,0), B(4,5) and C(-1,4), D(-2,-1). Then

AB = \(\sqrt{\left ( 3-4 \right )^{2}+\left ( 0-5 \right )^{2}}\)

= \(\sqrt{\left (-1\right )^{2}+\left ( -5 \right )^{2}}\)

= \(\sqrt{1+25}\)

= \(\sqrt{26}\) units

BC = \(\sqrt{\left (4+1 \right )^{2}+\left ( 5-4 \right )^{2}}\)

= \(\sqrt{\left (5\right )^{2}+\left ( 1 \right )^{2}}\)

= \(\sqrt{25+1}\)

= \(\sqrt{26}\) units

CD = \(\sqrt{\left ( -1+2 \right )^{2}+\left ( 4+1 \right )^{2}}\)

= \(\sqrt{\left (1\right )^{2}+\left ( 5 \right )^{2}}\)

= \(\sqrt{1+25}\)

= \(\sqrt{26}\) units

AD = \(\sqrt{\left ( 3+2 \right )^{2}+\left ( 0+1 \right )^{2}}\)

= \(\sqrt{\left (5\right )^{2}+\left ( 1 \right )^{2}}\)

= \(\sqrt{25+1}\)

= \(\sqrt{26}\) units

AC = \(\sqrt{\left ( 3+1 \right )^{2}+\left ( 0-4 \right )^{2}}\)

= \(\sqrt{\left (4\right )^{2}+\left ( -4 \right )^{2}}\)

= \(\sqrt{16+16}\)

= \(\sqrt{32}\)

= \(4\sqrt{2}\) units

BD = \(\sqrt{\left ( 4+2 \right )^{2}+\left ( 5+1 \right )^{2}}\)

= \(\sqrt{\left (6\right )^{2}+\left ( 6 \right )^{2}}\)

= \(\sqrt{36+36}\)

= \(\sqrt{72}\)

= \(6\sqrt{2}\) units

Therefore AB = BC = CD = DA = \(\sqrt{26}\) units and diagonal AC is not equal to diagonal BD

Hence, ABCD is a rhombus

Area of a rhombus = \(\frac{1}{2}*\left ( Product\;of\;its\;diagonals\right )\)

= \(\frac{1}{2}*4\sqrt{2}*6\sqrt{2}\)

= \(\frac{24*2}{2}\)

= 24 square units

**Q4) Show that the points A(6,1), B(8,2), C(9,4) and D(7,3) are the vertices of a rhombus. Find its area.**

The given points are A (6,1), B(8,2) and C(9,4), D(7,3). Then

AB = \(\sqrt{\left ( 6-8 \right )^{2}+\left ( 1-2 \right )^{2}}\)

= \(\sqrt{\left (-2\right )^{2}+\left ( -1 \right )^{2}}\)

= \(\sqrt{4+1}\)

= \(\sqrt{5}\) units

BC = \(\sqrt{\left (9-7 \right )^{2}+\left (4-3 \right )^{2}}\)

= \(\sqrt{\left (2\right )^{2}+\left ( 1 \right )^{2}}\)

= \(\sqrt{4+1}\)

= \(\sqrt{5}\) units

CD = \(\sqrt{\left ( 9-7 \right )^{2}+\left ( 4-3 \right )^{2}}\)

= \(\sqrt{\left (2\right )^{2}+\left ( 1 \right )^{2}}\)

= \(\sqrt{4+1}\)

= \(\sqrt{5}\) units

AD = \(\sqrt{\left ( 7-6 \right )^{2}+\left ( 3-1 \right )^{2}}\)

= \(\sqrt{\left (1\right )^{2}+\left ( 2 \right )^{2}}\)

= \(\sqrt{1+4}\)

= \(\sqrt{5}\) units

AC = \(\sqrt{\left ( 6-9 \right )^{2}+\left ( 1-4 \right )^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( -3 \right )^{2}}\)

= \(\sqrt{9+9}\)

= \(\sqrt{18}\)

= \(3\sqrt{2}\) units

BD = \(\sqrt{\left ( 8-7 \right )^{2}+\left ( 2-3 \right )^{2}}\)

= \(\sqrt{\left (1\right )^{2}+\left ( 1 \right )^{2}}\)

= \(\sqrt{1+1}\)

= \(\sqrt{2}\)

= \(\sqrt{2}\) units

Therefore AB = BC = CD = DA = \(\sqrt{5}\) units and diagonal AC is not equal to diagonal BD

Hence, ABCD is a rhombus

Area of a rhombus = \(\frac{1}{2}*\left ( Product\;of\;its\;diagonals\right )\)

= \(\frac{1}{2}*3\sqrt{2}*\sqrt{2}\)

= \(\frac{3*2}{2}\)

= 3 square units

**Q5) Show that the points A(2,1), B(5,2), C(6,4) and D(3,3) are the angular points of parallelogram. Is the figure a rectangle ?**

The given points are A (2,1), B(5,2) and C(6,4), D(3,3). Then

AB = \(\sqrt{\left ( 5-2 \right )^{2}+\left ( 2-1 \right )^{2}}\)

= \(\sqrt{\left (3\right )^{2}+\left ( 1 \right )^{2}}\)

= \(\sqrt{9+1}\)

= \(\sqrt{10}\) units

BC = \(\sqrt{\left (6-5 \right )^{2}+\left (4-2 \right )^{2}}\)

= \(\sqrt{\left (1\right )^{2}+\left ( 2 \right )^{2}}\)

= \(\sqrt{4+1}\)

= \(\sqrt{5}\) units

CD = \(\sqrt{\left ( 3-6 \right )^{2}+\left ( 3-4 \right )^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( -1 \right )^{2}}\)

= \(\sqrt{9+1}\)

= \(\sqrt{10}\) units

AD = \(\sqrt{\left ( 3-2 \right )^{2}+\left ( 3-1 \right )^{2}}\)

= \(\sqrt{\left (1\right )^{2}+\left ( 2 \right )^{2}}\)

= \(\sqrt{1+4}\)

= \(\sqrt{5}\) units

Thus AB = CD = \(\sqrt{10}\) units and BC = AD = \(\sqrt{5}\) units

So, quadrilateral ABCD is a parallelogram

Also, AC = \(\sqrt{\left ( 6-2 \right )^{2}+\left ( 4-1 \right )^{2}}\)

= \(\sqrt{\left (4\right )^{2}+\left ( 3 \right )^{2}}\)

= \(\sqrt{16+9}\)

= \(\sqrt{25}\)

= 5 units

BD = \(\sqrt{\left ( 3-5 \right )^{2}+\left ( 3-2 \right )^{2}}\)

= \(\sqrt{\left (-2\right )^{2}+\left ( 1 \right )^{2}}\)

= \(\sqrt{4+1}\)

= \(\sqrt{5}\)

= \(\sqrt{5}\) units

But diagonal AC is not equal to diagonal BD

Hence, the given points do not form a rectangle

**Q6) Show that A(1,2), B(4,3), C(6,6) and D(3,5) are the vertices of a parallelogram. Show that ABCD is not a rectangle.**

The given points are A (1,2), B(4,3) and C(6,6), D(3,5). Then

AB = \(\sqrt{\left (1-4 \right )^{2}+\left ( 2-3 \right )^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( -1 \right )^{2}}\)

= \(\sqrt{9+1}\)

= \(\sqrt{10}\) units

BC = \(\sqrt{\left (4-6 \right )^{2}+\left (3-6 \right )^{2}}\)

= \(\sqrt{\left (-2\right )^{2}+\left ( -3 \right )^{2}}\)

= \(\sqrt{4+9}\)

= \(\sqrt{13}\) units

CD = \(\sqrt{\left (6-3 \right )^{2}+\left ( 6-5 \right )^{2}}\)

= \(\sqrt{\left (3\right )^{2}+\left ( 1 \right )^{2}}\)

= \(\sqrt{9+1}\)

= \(\sqrt{10}\) units

AD = \(\sqrt{\left (1-3 \right )^{2}+\left (2-5 \right )^{2}}\)

= \(\sqrt{\left (-2\right )^{2}+\left ( -3 \right )^{2}}\)

= \(\sqrt{4+9}\)

= \(\sqrt{13}\) units

Thus AB = CD = \(\sqrt{10}\) units and BC = AD = \(\sqrt{13}\) unit

Therefore ABCD is a parallelogram

Also, AC = \(\sqrt{\left ( 1-6 \right )^{2}+\left ( 2-6 \right )^{2}}\)

= \(\sqrt{\left (-5\right )^{2}+\left ( -4 \right )^{2}}\)

= \(\sqrt{25+16}\)

= \(\sqrt{41}\)

= \(\sqrt{41}\) units

BD = \(\sqrt{\left ( 4-3 \right )^{2}+\left ( 3-5 \right )^{2}}\)

= \(\sqrt{\left (1\right )^{2}+\left ( -2 \right )^{2}}\)

= \(\sqrt{1+4}\)

= \(\sqrt{5}\)

= \(\sqrt{5}\) units

But diagonal AC is not equal to diagonal BD

Hence, ABCD is not a rectangle

**Q7) Show that the following points are the vertices of a rectangle.**

**(i) A (-4,-1), B(-2,-4) and C(4,0), D(2,3).**

**(ii) A (2,-2), B(14,10) and C(11,13), D(-1,1).**

**(iii) A (0,-4), B(6,2) and C(3,5), D(-3,-1).**

The given points are A (-4,-1), B(-2,-4) and C(4,0), D(2,3). Then

AB = \(\sqrt{\left \{ -2-\left ( -4 \right )\right \}^{2}+\left \{ -4-\left ( -1 \right ) \right \}^{2}}\)

= \(\sqrt{\left (2\right )^{2}+\left ( -3 \right )^{2}}\)

= \(\sqrt{4+9}\)

= \(\sqrt{13}\)

= \(\sqrt{13}\) units

BC = \(\sqrt{\left \{ 4-\left ( -2 \right )\right \}^{2}+\left \{ 0-\left ( -4 \right ) \right \}^{2}}\)

= \(\sqrt{\left (6\right )^{2}+\left ( 4 \right )^{2}}\)

= \(\sqrt{36+16}\)

= \(\sqrt{52}\)

= \(2\sqrt{13}\) units

CD = \(\sqrt{\left ( 2-4 \right )^{2}+\left ( 3-0 \right )^{2}}\)

= \(\sqrt{\left (-2\right )^{2}+\left ( 3 \right )^{2}}\)

= \(\sqrt{4+9}\)

= \(\sqrt{13}\)

= \(\sqrt{13}\) units

AD = \(\sqrt{\left \{ 2-\left ( -4 \right )\right \}^{2}+\left \{ 3-\left ( -1 \right ) \right \}^{2}}\)

= \(\sqrt{\left (6\right )^{2}+\left ( 4 \right )^{2}}\)

= \(\sqrt{36+16}\)

= \(\sqrt{52}\)

= \(2\sqrt{13}\) units

Thus AB = CD = \(\sqrt{13}\) units and BC = AD = \(2\sqrt{13}\) unit

Also, AC = \(\sqrt{\left \{ 4-\left ( -4 \right )\right \}^{2}+\left \{ 0-\left ( -1 \right ) \right \}^{2}}\)

= \(\sqrt{\left (8\right )^{2}+\left ( 1 \right )^{2}}\)

= \(\sqrt{64+1}\)

= \(\sqrt{65}\)

= \(2\sqrt{65}\) units

BD = \(\sqrt{\left \{ 2-\left ( -2 \right )\right \}^{2}+\left \{ 3-\left ( -4 \right ) \right \}^{2}}\)

= \(\sqrt{\left (4\right )^{2}+\left ( 7 \right )^{2}}\)

= \(\sqrt{16+49}\)

= \(\sqrt{65}\)

= \(2\sqrt{65}\) units

Also, diagonal AC = diagonal BD

Hence, the given points form a rectangle.

**ii) A (2,-2), B(14,10) and C(11,13), D(-1,1).**

The given points are A (2,-2), B(14,10) and C(11,13), D(-1,1). Then

AB = \(\sqrt{\left ( 14-2 \right )^{2}+ \left \{ 10-\left ( -2 \right ) \right \}^{2}}\)

= \(\sqrt{\left (12\right )^{2}+\left ( 23 \right )^{2}}\)

= \(\sqrt{144+144}\)

= \(\sqrt{288}\)

= \(12\sqrt{2}\) units

BC = \(\sqrt{\left ( 11-14 \right )^{2}+\left ( 13-10 \right )^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( 3 \right )^{2}}\)

= \(\sqrt{9+9}\)

= \(\sqrt{18}\)

= \(3\sqrt{2}\) units

CD = \(\sqrt{\left ( -1-11 \right )^{2}+\left ( 1-13 \right )^{2}}\)

= \(\sqrt{\left (-12\right )^{2}+\left ( -12 \right )^{2}}\)

= \(\sqrt{144+144}\)

= \(\sqrt{288}\)

= \(12\sqrt{2}\) units

AD = \(\sqrt{\left ( -1-2 \right )^{2}+ \left \{ 1-\left ( -2 \right ) \right \}^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( 3 \right )^{2}}\)

= \(\sqrt{9+9}\)

= \(\sqrt{18}\)

= \(3\sqrt{2}\) units

Thus AB = CD = \(12\sqrt{2}\) units and BC = AD = \(3\sqrt{2}\) unit

Also, AC = \(\sqrt{\left ( 11-2 \right )^{2}+ \left \{ 13-\left ( -2 \right ) \right \}^{2}}\)

= \(\sqrt{\left (9\right )^{2}+\left ( 15 \right )^{2}}\)

= \(\sqrt{81+225}\)

= \(\sqrt{306}\)

= \(3\sqrt{34}\) units

BD = \(\sqrt{\left ( -1-14 \right )^{2}+\left ( 1-10 \right )^{2}}\)

= \(\sqrt{\left (-15\right )^{2}+\left ( -9 \right )^{2}}\)

= \(\sqrt{81+225}\)

= \(\sqrt{306}\)

= \(3\sqrt{34}\) units

Also, diagonal AC = diagonal BD

Hence, the given points form a rectangle.

**iii) A (0,-4), B(6,2) and C(3,5), D(-3,-1).**

The given points are A (0,-4), B(6,2) and C(3,5), D(-3,-1). Then

AB = \(\sqrt{\left ( 6-0 \right )^{2}+ \left \{ 2-\left ( -4 \right ) \right \}^{2}}\)

= \(\sqrt{\left (6\right )^{2}+\left ( 6 \right )^{2}}\)

= \(\sqrt{36+36}\)

= \(\sqrt{72}\)

= \(6\sqrt{2}\) units

BC **= ** \(\sqrt{\left ( 3-6 \right )^{2}+\left ( 5-2 \right )^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( 3 \right )^{2}}\)

= \(\sqrt{9+9}\)

= \(\sqrt{18}\)

= \(3\sqrt{2}\) units

CD = \(\sqrt{\left ( -3-3\right )^{2}+\left ( -1-5 \right )^{2}}\)

= \(\sqrt{\left (-6\right )^{2}+\left ( -6 \right )^{2}}\)

= \(\sqrt{36+36}\)

= \(\sqrt{72}\)

= \(6\sqrt{2}\) units

AD = \(\sqrt{\left ( -3-0 \right )^{2}+ \left \{ -1-\left ( -4 \right ) \right \}^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( 3 \right )^{2}}\)

= \(\sqrt{9+9}\)

= \(\sqrt{18}\)

= \(3\sqrt{2}\) units

Thus AB = CD = \(6\sqrt{2}\) units and BC = AD = \(3\sqrt{2}\) units

Also, AC = \(\sqrt{\left ( 3-0 \right )^{2}+ \left \{ 5-\left ( -4 \right ) \right \}^{2}}\)

= \(\sqrt{\left (3\right )^{2}+\left ( 9 \right )^{2}}\)

= \(\sqrt{9+81}\)

= \(\sqrt{90}\)

= \(3\sqrt{10}\) units

BD = \(\sqrt{\left ( -3-6 \right )^{2}+\left ( -1-2 \right )^{2}}\)

= \(\sqrt{\left (-9\right )^{2}+\left ( -3 \right )^{2}}\)

= \(\sqrt{81+9}\)

= \(\sqrt{90}\)

= \(3\sqrt{10}\) units

Also, diagonal AC = diagonal BD

Hence, the given points

form a rectangle.