Solution:

A(−4, 0) = (x1, y1)
B(0,−3) = (x2, y2)
AB = ?
We know that,
or, AB = \(\sqrt{(x_2− x_1)^2 + (y_2−y_1)^2}\)
or, AB = \(\sqrt{(0 + 4)^2 + (− 3 − 0)^2}\)
or, AB = \(\sqrt{4)^2 + (3)^2}\)
or, AB = \(\sqrt{16 + 9}\)
or, AB = \(\sqrt{25}\)
∴ AB = 5 units.



Solution:

Here,
P(−8, 0) = (x1, y1)
B(0, 6) =(x2, y2)
PB = ?
We know that,
or, PB = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, PB = \(\sqrt{(0 + 8)^2 + (6 − 0)^2}\)
or, PB = \(\sqrt{(8)^2 + (6)^2}\)
or, PB = \(\sqrt{64 + 36}\)
or, PB = \(\sqrt{100}\)
∴ PB = 10 units.

Solution:

Here,
A(0,−5) = (x1, y1)
B(4 , 8).= (x2, y2)
Slope (m) = ?

We know that,
Slope (m) = \(\frac{y_2− y_1}{x_2− x_1}\)
=\(\frac{8 + 5}{4 − 0}\)
= \(\frac{13}{4}\)

Solution:

Here,
E(4,−7) = (x2, y1)
F(−3, 4) = (x2, y2)
Slope (m) = ?

we know that,
Slope (m) = \(\frac{y_2− y_1}{x_2− x_1}\)
=\(\frac{4 + 7}{−3 − 4}\)
= \(\frac{11}{−7}\)

Solution:

Let,
A (−1, 7) = (x1, y1)
B (3, 10) = (x2, y2)
Distance of AB = ?

We know that,
or, AB = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, AB = \(\sqrt{(3 + 1)^2 + (10 − 7)^2}\)
or, AB = \(\sqrt{(4)^2 + (3)^2}\)
or, AB = \(\sqrt{16+ 9}\)
or, AB = \(\sqrt{25}\)
∴ AB = 5 units

Solution:

Let, Q(0, 0) and R(−6,−4)
Distance of QR = ?
We know that,
or, QR = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, QR = \(\sqrt{(− 6 − 0)^2 + (− 4 − 0)^2}\)
or, QR = \(\sqrt{(6)^2 + (4)^2}\)
or, QR = \(\sqrt{36 + 16}\)
or, QR = \(\sqrt{52}\)
∴ QR = 2\(\sqrt{3}\)

Solution:

Here,
A(4, 2) = (x1, y1)
B(6, 8) = (x2, y2)
Slope (m) = ?
We know that,
Slope (m) = \(\frac{y_2− y_1}{x_2− x_1}\)
=\(\frac{8 − 2}{6 − 4}\)
=\(\frac{6}{2}\)
= 3

Solution:

For, AB
A(2, 3) = (x, y)
B(2, 0) = (x2, y2)
AB = ?
We know that,
or, AB = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, AB =\(\sqrt{(2− 2)^2 + (0 − 3)^2}\)
or, AB =\(\sqrt{(0)^2 + (3)^2}\)
or, AB =\(\sqrt{3^2}\)
∴ AB = 3 units.

For BC
B(2, 0) = (x1, y1)
C(−1, 0) = (x2, y2)
BC = ?
We know that,
or, BC = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, BC = \(\sqrt{(−1 − 2)^2 + 0 − 0)^2}\)
or, BC = \(\sqrt{(3)^2}\)
∴ BC = 3 units

For CA
C(− 1, 0) = (x1, y1)
A(2, 3) =(x2, y2)
CA = ?
We know that,
or, CA =\(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, CA =\(\sqrt{(2 + 1)^2 + (3 − 0)^2}\)
or, CA =\(\sqrt{(3)^2 + (3)^2}\)
or, CA =\(\sqrt{9 + 9}\)
or, CA =\(\sqrt{18}\)
∴ 3\(\sqrt{2}\) units

Solution:

For AB
A(−3,−4) = (x1, y1)
B(−1,−2) =(x2, y2)
AB = ?
We know that,
or, AB = \(\sqrt{(x_2−x_1)^2 + (y_2− y_1)^2}\)
or, AB = \(\sqrt{(−1 + 3)^2 + (−2 + 4)^2}\)
or, AB = \(\sqrt{(2)^2 + (2)^2}\)
or, AB = \(\sqrt{4 +4}\)
or, AB = \(\sqrt{8}\)
or, AB = 2\(\sqrt{2}\) units.

For CD
C(8, 6) = (x1, y1)
D(10, 8) =(x2, y2)
CD = ?
We know that,
or, CD = \(\sqrt{(x_2−x_1)^2 + (y_2− y_1)^2}\)
or, CD = \(\sqrt{(10 − 8)^2 + (8 − 6)^2}\)
or, CD = \(\sqrt{(2)^2 + (2)^2}\)
or, CD = \(\sqrt{4+4}\)
or, CD =\(\sqrt{8}\)
or, CD = 2\(\sqrt{2}\) units.

∴ AB = CD =2\(\sqrt{2}\) units. proved.



Solution:

For PQ
P(2,−5) = (x1, y1)
Q(4, 1) = (x2, y2)
PQ = ?
We know that,
or, PQ = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, PQ = \(\sqrt{(4 − 2)^2 + (1 +5)^2}\)
or, PQ = \(\sqrt{(2)^2 + (6)^2}\)
or, PQ = \(\sqrt{4 + 36}\)
or, PQ = \(\sqrt{40}\)
∴ PQ = 2\(\sqrt{10}\) units.

For QR
Q(4, 1) =(x1, y1)
R(10, 3) =(x2, y2)
QR = ?
We know that,
or, QR = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, QR = \(\sqrt{(10− 4)^2 + (3 − 1)^2}\)
or, QR = \(\sqrt{(6)^2 + (2)^2}\)
or, QR = \(\sqrt{36 + 4}\)
or, QR = \(\sqrt{40}\)
∴ QR = 2\(\sqrt{10}\)
∴ PQ = QR proved.

Solution:

Let, A(1, 2), B(4, 5) and C(7, 2) be the three points
For AB
A(1, 2) = (x1, y1)
B(4, 5) = (x2, y2)
Distance of AB = ?
We knoe that,
or, AB = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, AB = \(\sqrt{(4 − 1)^2 + (5 − 2)^2}\)
or, AB = \(\sqrt{(3)^2 + (3)^2}\)
or, AB = \(\sqrt{18}\)
or, AB = 3\(\sqrt{2}\) units

For BC
B(4, 5) = (x1, y1)
C(7, 2) = (x2, y2)
BC = ?
we know that,
or, BC = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, BC = \(\sqrt{(7− 4)^2 + (2 − 5)^2}\)
or,BC = \(\sqrt{(3)^2 + (− 3)^2}\)
or, BC = \(\sqrt{9 + 9}\)
or, BC = 3\(\sqrt{2}\) units

For CA
C(7, 2) = (x1, y1)
A(1, 2) = (x2, y2)
CA = ?
We know that,
or, CA = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, CA = \(\sqrt{(1 − 7)^2 + (2− 2)^2}\)
or, CA = \(\sqrt{( − 6)^2 + (0)^2}\)
or, CA = \(\sqrt{36}\)
or, CA = 6 units

Now,
CA has the longest side
or, 62 = (3\(\sqrt{2}\))2+(3\(\sqrt{2}\))2
or, 36 = 18 + 18
or, 36 = 36

AB = BC = 3\(\sqrt{2}\) units and is satisfied the pythogorus theorem (i.e. h2 = p2 + b2). So, the given points are the vertices of isosceles right angled triangle

Solution:

For AB
A(1,−1) = (x1, y1)
B(−1, 1) = (x2, y2)
AB = ?
We know that,
or, AB = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, AB = \(\sqrt{(− 1− 1)^2 + (1 +1)^2}\)
or, AB = \(\sqrt{(−2)^2 + 2^2}\)
or, AB = \(\sqrt{4 + 4}\)
or, AB = \(\sqrt{8}\)
or, AB = 2\(\sqrt{2}\) units.

For BC
B(−1, 1) = (x1, y1)
C(\(\sqrt{3}\), \(\sqrt{3}\)) = (x2, y2)
BC = ?
We know that,
or, BC = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, BC = \(\sqrt{(\sqrt{3}+1)^2 + (\sqrt{3} − 1)^2}\)
or, BC = \(\sqrt{(\sqrt{3})^2 + 2 . \sqrt{3} .1 + 1^2 + (\sqrt{3})^2 − 2 . \sqrt{3} . 1 + 1^2}\)
or, BC = \(\sqrt{3 +2 . \sqrt{3} + 1 + 3− 2 . \sqrt{3} + 1}\)
or, BC = \(\sqrt{8}\)
or, BC = 2\(\sqrt{2}\)

For AC
A(1,−1) = (x1, y1)
C(\(\sqrt{3}\),\(\sqrt{3}\)) = (x2, y2)
AC = ?
We know that,
or, AC = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, AC = \(\sqrt{(\sqrt{3} − 1)^2 + (\sqrt{3}+1)^2}\)
or, AC = \(\sqrt{(\sqrt{3})^2 − 2 . \sqrt{3} .1 + 1^2 + (\sqrt{3})^2 +2 . \sqrt{3} . 1 + 1^2}\)
or, AC = \(\sqrt{3 − 2 . \sqrt{3} + 1 + 3 +2 . \sqrt{3} + 1}\)
or, AC = \(\sqrt{8}\)
or, AC = 2\(\sqrt{2}\)

Here, AB = BC = AC. So, the given points are the vertices of an equilateral triangle.

Solution:

For AB
A(1, 1) = (x1, y1)
B(4, 4) = (x2, y2)
AB = ?
We know that,
or, AB = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or,AB = \(\sqrt{(4 − 1)^2 + (4 − 1)^2}\)
or,AB = \(\sqrt{(3)^2 + (3)^2}\)
or, AB = \(\sqrt{9 + 9}\)
or, AB = \(\sqrt{18}\)
or, AB = 3\(\sqrt{2}\) units

For BC
B(4, 4) =(x1, y1)
C(4, 8) = (x2, y2)
BC = ?
We know that,
or, BC =\(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, BC =\(\sqrt{(4− 4)^2 + (8− 4)^2}\)
or, BC = \(\sqrt{4^2}\)
or, BC = 4 units.

For CD
C(4, 8) =(x1, y1)
D(1, 5) =(x2, y2)
CD = ?
We know that,
or, CD =\(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, CD =\(\sqrt{(1 − 4)^2 + (5 − 8)^2}\)
or, CD =\(\sqrt{(− 3)^2 + (− 3)^2}\)
or, CD = \(\sqrt{9 + 9}\)
or, CD = 3\(\sqrt{2}\) units

For AD
A(1, 1) =(x1, y1)
D(1, 5) =(x2, y2)
AD = ?
We know that,
or, AD =\(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, AD =\(\sqrt{(1− 1)^2 + (5 − 1)^2}\)
or, AD = \(\sqrt{4^2}\)
or, AD = 4 units

Here, AB = CD = 3\(\sqrt{2}\) units
BC = AB = 4 units
So, the given points are the vertices of a parallelogram.

Solution:

For AB
A(0,−1) = (x1, y1)
B(2, 1) = (x2, y2)
AB = ?
We know that,
or, AB = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, AB = \(\sqrt{2− 0)^2 + (1+1)^2}\)
or, AB = \(\sqrt{2^2 + 2^2}\)
or, AB = \(\sqrt{4 + 4}\)
or, AB = \(\sqrt{8}\)
or, AB = 2\(\sqrt{2}\) units

For BC
B(2, 1) = (x1, y1)
C(0, 3) = (x2, y2)
BC = ?
We know that,
or, BC = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, BC = \(\sqrt{(0− 2)^2 + (3 − 1)^2}\)
or, BC = \(\sqrt{(− 2)^2 + (2)^2}\)
or, BC = \(\sqrt{4 + 4}\)
or, BC = \(\sqrt{8}\)
or, BC = 2\(\sqrt{2}\) units

For CD
C(0, 3) = (x1, y1)
D(−2, 1) = (x2, y2)
CD = ?
We know that,
or, CD = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, CD = \(\sqrt{(− 2− 0)^2 + (1 − 3)^2}\)
or, CD = \(\sqrt{(− 2)^2 + 2^2}\)
or, CD = \(\sqrt{4 + 4}\)
or, CD = \(\sqrt{8}\)
or, CD = 2\(\sqrt{2}\) units

For AD
A(0,−1) = (x1, y1)
D(−2, 1) = (x2, y2)
AD = ?
We know that,
or, AD = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, AD = \(\sqrt{(− 2− 0)^2 + (1 − 1)^2}\)
or, AD = \(\sqrt{(− 2)^2 + (2)^2}\)
or, AD = \(\sqrt{4 + 4}\)
or, AD = \(\sqrt{8}\)
or, AD = 2\(\sqrt{2}\) units

For AC
A(0,−1) = (x1, y1)
C(0, 3) = (x2, y2)
or, AC = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, AC = \(\sqrt{(0 − 0)^2 + (3+1)^2}\)
or, AC = \(\sqrt{4^2}\)
or, AC = 4 units.

For DB
D(−2, 1) = (x1, y1)
B(2, 1) = (x2, y2)
DB = ?
We know that,
or, DB = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, DB = \(\sqrt{(2+2)^2 + (1 − 1)^2}\)
or, DB = \(\sqrt{4^2}\)
or, DB = 4 units

Here, AB = BC = CD = AD = 2\(\sqrt{[2}\) units
AC = BD = 4 units
Since the given points are the vertices of a square.

Solution:

Let, A(2,−1), B(7, 4) and C(8, 11) be the vertices of an isoscles trinagle.
For AB
A(2,−1) = (x1, y1)
B(7, 4) = (x2, y2)
AB = ?
We know that,
or, AB = \(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, AB =\(\sqrt{(7 − 2)^2 + (4 +1)^2}\)
or, AB =\(\sqrt{(5)^2 + (5)^2}\)
or, AB = \(\sqrt{25 + 25}\)
or, AB = \(\sqrt{50}\)
or, AB = 5\(\sqrt{2}\) units

For BC
B(7, 4) =(x1, y1)
C(8, 11) =(x2, y2)
BC = ?
we know that,
or, BC =\(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, BC =\(\sqrt{(8 − 7)^2 + (11 − 4)^2}\)
or, BC = \(\sqrt{1 + 7^2 }\)
or, BC = \(\sqrt{50}\)
or, BC = 5\(\sqrt{2}\) units

For AC
A(2,−1) = (x1, y1)
C(8, 11) =(x2, y2)
AC = ?
We know that,
or, AC =\(\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}\)
or, AC =\(\sqrt{(8 − 2)^2 + (11+1)^2}\)
or, AC =\(\sqrt{(6)^2 + (12)^2}\)
or, AC = \(\sqrt{36 + 144}\)
or, AC = \(\sqrt{180}\)
or, AC = 6\(\sqrt{5}\) units

Here, AB = BC = 5\(\sqrt{2}\) units
So, the given points are the vertices of an isosceles triangle

Solution:

For AB
A((1,?1) = (x1, y1)
B(3, 2) = (x2, y2)
AB = ?
We know that,
or, AB = \(\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}\)
or, AB = \(\sqrt{(3 ? 1)^2 + (2 ? 1)^2}\)
or, AB = \(\sqrt{(2)^2 + (3)^2}\)
or, AB = \(\sqrt{4 + 9}\)
or, AB = \(\sqrt{13}\) units

For BC
B(3, 2) = (x1, y1)
C(1, 4) = (x2, y2)
BC = ?
We know that,
or, BC = \(\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}\)
or, BC = \(\sqrt{(1 ? 3)^2 + (4 ? 2)^2}\)
or, BC = \(\sqrt{(? 2)^2 + (2)^2}\)
or, BC = \(\sqrt{4 + 4}\)
or, BC = \(\sqrt{8}\) units

For CD
C(1, 4) = (x1, y1)
D(?2, 2) = (x2, y2)
CD = ?
We know that,
or, CD = \(\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}\)
or, CD = \(\sqrt{(? 2 ? 1)^2 + (2 ? 4)^2}\)
or, CD = \(\sqrt{(? 3)^2 + (? 2)^2}\)
or, CD = \(\sqrt{9 + 4}\)
or, CD = \(\sqrt{13}\) units

For AD
A(1,?1) = (x1, y1)
D(?2, 2) = (x2, y2)
AD = ?
We know that,
or, AD = \(\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}\)
or, AD = \(\sqrt{(1 ? 1)^2 + (4+1)^2}\)
or, AD = \(\sqrt{(0)^2 + (5)^2}\)
or, AD = 5 units

For BC
B(3, 2) = (x1, y1)
D(?2, 2) = (x2, y2)
BD = ?
We know thats,
or, BD = \(\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}\)
or, BD = \(\sqrt{(? 2 ? 3)^2 + (2? 2)^2}\)
or, BD = \(\sqrt{(?5)^2 + (0)^2}\)
or, BD = 5 units

Here,
AC = BD = 5 units
But other points cannot be proved. So it is not rectangle.