# RS Aggarwal Solutions Class 10 Ex 16B

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.

#### Practise This Question

Mendeleev’s periodic law is based on