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(i) (2,3), (4,1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (-a, -b)
Solution:
Formula used in the above question is : √(x2 - x1)2 + (y2 - y1)2 (i.e., Distance Formula)
(i) Here, x1 = 2, y1= 3, x2 = 4, y2 = 1
Now, applying the distance formula :
= √(4-2)2 + (1-3)2
=√(2)2 + (-2)2
= √8
= 2√2 units
(ii) Here, x1= -5, y1= 7, x2 = -1, y2 = 3
Now, applying the distance formula :
= √(-1 - (-5))2 + (3 - 7)2
= √(4)2 + (-4)2
= 4√2
(iii) Here, x1 = a, y1 = b . x2 = -a, y2 = -b
Now, applying the distance formula :
= √(-a - a)2 +(-b - b)2
= √(-2a)2 + (-2b)2
= √4a2 + 4b2
= 2√a2 + b2
Solution:
Formula used in the above question is : √(x2 - x1)2 + (y2 - y1)2 (i.e, Distance Formula)
Considering, Point A as (0, 0) and Point B as (36, 15) and applying the distance formula we get :
Distance between the two points : √(36 - 0)2 + (15 - 0)2
= √(36)2 + (15)2
= √1296 + 225
= √1521
= 39 units
Hence, the distance between two towns A and B is 39 units.
Solution:
Let (1, 5), (2, 3) and (-2, -11) be points A, B and C respectively.
Collinear term means that these 3 points lie in the same line. So, to we' ll check it.
Using distance formula we will find the distance between these points.
AB = √(2 - 1)2 + (3 - 5)2
=√(1)2 + (-2)2 =√1 + 4 =√5
BC = √(-2 - 2)2 + (-11 - 3)2
= √(-4)2 + (-14)2 = √16 + 196 = √212
CA = √(-2 - 1)2 + (-11 - 5)2
= √(-3)2 + (-16)2 = √9 + 256 =√265
As, AB + BC ≠ AC (Since, one distance is not equal to sum of other two distances, we can say that they do not lie in the same line.)
Hence, points A, B and C are not collinear.
Solution:
Let (5, – 2), (6, 4) and (7, – 2) be the points A, B and C respectively.
Using distance formula :
AB = √(6 - 5)2 + (4 - (-2))2
= √(1 + 36) = √37
BC = √(7 - 6)2 + (-2 - 4)2
= √(1 + 36) = √37
AC = √(7 - 5)2 + (-2 - (-2))2
= √(4 + 0) = 2
As, AB = BC ≠ AC (Two distances equal and one distance is not equal to sum of other two)
So, we can say that they are vertices of an isosceles triangle.
Solution:
From the given fig, find the coordinates of the points
AB = √(6 - 3) + (7 - 4)
= √9+9 = √18 = 3√2
BC = √(9 - 6) + (4 - 7)
= √9+9 = √18 = 3√2
CD = √(6 - 9) + (1 - 4)
= √9 + 9 = √18 = 3√2
DA = √(6 - 3) + (1 - 4)
= √9+9 =√18 =3√2
AB = BC = CD = DA = 3√2
All sides are of equal length. Therefore, ABCD is a square and hence, Champa was correct.
(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
(i) Here, let the given points are P(-1, -2), Q(1, 0), R(-1, 2) and S(-3, 0) respectively.
PQ = √(1 - (-1))² + (0 - (-2))²
= √(1+1)²+(0+2)²
= √8 = 2 √2
QR = √(−1−1)²+(2−0)²
= √(−2)²+(2)²
= √8 = 2 √2
RS = √(−3−(−1))²+(0−2)²
= √8 = 2 √2
PS = √((−3−(−1))²+(0−(−2))²
= √8 = 2 √2
Here, we found that the length of all the sides are equal.
Diagonal PR = √(−1−(−1))²+(2−(−2))²
= √ 0+16
= 4
Diagonal QS = √(−3−1)²+(0−0)²
= √ 16 = 4
Finally, we also found that the length of diagonal are also same.
Here, PQ = QR = RS = PS = 2√2
and QS = PR = 4
This is the property of SQUARE. Hence, the given figure is SQUARE.
(ii) Let the points be P(-3, 5), Q(3, 1), R(0, 3) and S(-1, -4)
PQ = √(3−(−3))²+(1−5)²
= √(3+3)²+(−4)²
= √36+16 = √52 = 2 √13
QR = √(0−3)²+(3−1)²
= √(−3)²+(2)²
= √9 + 4 = √13
RS = √(−1−0)²+(−4−3)²
= √(−1)²+(−7)²
= √1+49 = √50 = 5 √2
PS = √(−1−(−3))²+(−4−5)²
= √(−1+3)²+(−9)²
= √4+81 = √85
Here, All the lengths of sides are unequal.
So, The given points will not create any quadrilateral.
(iii) Let the points be P(4, 5), Q(7, 6), R(4, 3) and S(1, 2)
PQ = √(7−4)²+(6−5)²
= √(3)²+(1)² = √9+1 = √10
QR = √(4−7)²+(3−6)²
= √(−3)²+(−3)² = √9+9 =3 √2
RS = √(1−4)²+(2−3)²
= √(−3)²+(−1)² = √9+1 = √10
PS = √(1−4)²+(2−5)²
= √(−3)²+(−3)² = √9+9 =3 √2
We see that the opposite sides are equal. Lets find the diagonal now.
Diagonal PR = √(4−4)²+(3−5)²
= √0+4 = 2
Diagonal QS = √(1−7)²+(2−6)²
= √36+16
= √52
Here, PQ = RS = √10
and QR = PS = 3√2
We see that the diagonals are not equal.
Hence, the formed quadrilateral is a PARALLELOGRAM.
Solution:
Let the point on the X axis be (x, 0)
Given, distance between the points (2, -5), (x, 0) = distance between points (-2, 9), (x, 0)
[ Applying Distance Formula ]
⇒ √(x - 2)²+(0 - (-5))² = √(x - (-2))²+(0 - 9)²
On squaring both the sides, we get
⇒ (x – 2)² + 5² = (x + 2)² + 9²
⇒ x² – 4x + 4 + 25 = x² + 4x + 4 + 81
⇒ -4x -4x = 85 – 29
⇒ -8x = 56
⇒ x = -7
Solution:
It is given that, the distance b/w two points is 10 units.
So, we'll find the distance and equate
PQ = √ (10 - 2)2 + (y - (-3))2
= √ (8)2 + (y +3)2
On squaring both the sides, we get :
64 +(y+3)2 = (10)2
(y+3)2 = 36
y + 3 = ±6
y + 3 = +6 or y + 3 = −6
Hence, y = 3 or -9.
Solution:
Given, PQ = QR
We will apply distance formula and find the distance between them,
PQ = √(5 - 0)2 + (-3 - 1)2
= √ (- 5)2 + (-4)2
= √ 25 + 16 = √41
QR = √ (0 - x)2 + (1 - 6)2
= √ (-x)2 + (-5)2
= √ x2 + 25
As they both are equal so on equating them, x2 + 25 = 41
x2 =16, x = ± 4
So, putting the value of x and obtaining the value of QR and PR through distance formula
For x = +4, PR = √ (4 - 5)2 + (6 - (-3))2
= √ (-1) 2+ (9)2
= √ 82
QR = √ (0 - 4)2 + (1 - 6)2
= √ 41
For x = -4, QR = √ (0 - (-4))2 + (1 - 6)2
= √ 16 + 25 = √ 41
PR = √ (5 + 4)2 + (-3 -6)2
= √ 81 + 81 = 9 √ 2
Solution:
Let, (x, y) be point P and (3, 6), (-3, 4) be pt A and B respectively.
It is given that their distance is equal, so we will equate the equations.
PA = PB (given)
⇒ √(x - 3)2 +(y - 6)2 = √(x-(-3))2+ (y - 4)2 [ By applying distance formula ]
On squaring both sides,
(x-3)2+(y-6)2 = (x +3)2 +(y-4)2
x2 +9-6x+y2+36-12y = x2 +9+6x+y2 +16-8y
36-16 = 6x+6x+12y-8y
20 = 12x+4y
3x+y = 5
3x+y-5 = 0
