(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (- a, – b)
(i) (2, 3), (4, 1)
Distance between two points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is given by
$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
Therefore distance between two points(2,3) and (4,1) is given by
=$\sqrt{(4-2)^{2}+(1-3)^{2}}$
=$\sqrt{(2)^{2}+(-2)^{2}}$
=$\sqrt{4+4}$
=$\sqrt{8}$
=$2\sqrt{2}$
(ii) (-5, 7), (-1, 3)
Distance between two points(-5,7) and (-1,3) is given by
=$\sqrt{(-1)-(-5))^{2}+(3-7)^{2}}$
=$\sqrt{(4))^{2}+(-4)^{2}}$
=$\sqrt{16+16}$
=$\sqrt{32}$
=$4\sqrt{2}$
(iii) (a, b), (- a, – b)
Distance between two points(a,b) and (-a,-b) is given by
=$\sqrt{(-a-a))^{2}+(-b-b)^{2}}$
=$\sqrt{(-2a)^{2}+(-2b)^{2}}$
=$\sqrt{(4a)^{2}+(4b)^{2}}$
=$2\sqrt{a^{2}+b^{2}}$
Let us consider, town A at point (0, 0). Therefore, town B will be at point (36, 15).
Distance between points(0,0) and (36,15)
Now Distance between two points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is given by
$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
=$\sqrt{(36-0)^{2}+(15-0)^{2}}$
=$\sqrt{(36)^{2}+(15)^{2}}$
=$\sqrt{1296+225}$
=$\sqrt{1521}$
=39
In section 7.2, A is (4,0) and B is (6,0).
$AB^{2}$ = $(6 - 4)^{2} - (0 - 0)^{2}$ = 4
The distance between town A and B will be 39 km.
The distance between the two towns A and B discussed in Section 7.2 is 4 km.
The sum of the lengths of any two line segments is equal to the length of the third line segment then all three points are collinear.
Consider, A = (1, 5) B = (2, 3) and C = (-2, -11)
Find the distance between points; say AB, BC and CA
AB=$\sqrt{(2-1)^{2}+(3-5)^{2}}$=$\sqrt{5}$
BC=$\sqrt{(-2-2)^{2}+(-11-3)^{2}}$
=$\sqrt{(-4)^{2}+(-14)^{2}}$
= $ \sqrt{16+196}$
= $\sqrt{212}$
CA=$\sqrt{(-2-1)^{2}+(-11-5)^{2}}$
= $\sqrt{(-3)^{2}+(-16)^{2}}$
= $\sqrt{9+256}$
= $ \sqrt{265}$
Since AB + BC ≠ CA
Therefore, the points (1, 5), (2, 3), and ( – 2, – 11) are not collinear.
Since two sides of any isosceles triangle are equal.
To check whether given points are vertices of an isosceles triangle, we will find the distance between all the points.
Let the points (5, – 2), (6, 4), and (7, – 2) are representing the vertices A, B, and C respectively.
AB=$\sqrt{(6-5)^{2}+(4+2)^{2}}$
=$ \sqrt{(-1)^{2}+(6)^{2}}$
=$ \sqrt{1+36}$
= $ \sqrt{37}$
BC=$\sqrt{(7-6)^{2}+(-2-4)^{2}}$
= $ \sqrt{(-1)^{2}+(6)^{2}}$
= $\sqrt{1+36}$
=$ \sqrt{37}$
CA=$\sqrt{(7-5)^{2}+(-2+2)^{2}}$
=$\sqrt{(-2)^{2}+(0)^{2}}$
=2
Here AB=BC=$ \sqrt{37}$
This implies, whether given points are vertices of an isosceles triangle.
From figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6,1).
Find distance between points using distance formula, we get
AB= $\sqrt{(6-3)^{2}+(7-4)^{2}}$
$ = \sqrt{(-3)^{2}+(-3)^{2}}$
$ = \sqrt{9+9}$
= $ \sqrt{18}$
$= 3\sqrt{2}$
BC= $\sqrt{(9-6)^{2}+(4-7)^{2}}$
$= \sqrt{(3)^{2}+(-3)^{2}}$
$ = \sqrt{9+9}$
$= \sqrt{18}$
$= 3\sqrt{2}$
CD=$\sqrt{(6-9)^{2}+(1-4)^{2}}$
$ = \sqrt{(-3)^{2}+(-3)^{2}}$
$ = \sqrt{9+9}$
$ = \sqrt{18}$
$3\sqrt{2}$
AD=$ \sqrt{(6-3)^{2}+(1-4)^{2}}$
$ = \sqrt{(3)^{2}+(-3)^{2}}$
$ = \sqrt{9+9}$
$= \sqrt{18}$
$= 3\sqrt{2}$
Diagonal AC= $\sqrt{(3-9)^{2}+(4-4)^{2}}$
$ \sqrt{(-6)^{2}+(0)^{2}}$
=6
Diagonal BD=$\sqrt{(6-6)^{2}+(7-1)^{2}}$
$ \sqrt{(0)^{2}+(6)^{2}}$
=6
Here , all sides of this square are of same length and also diagonals are of same length.
So,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)
(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)
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= $\sqrt{(1+1)^{2}+(0+2)^{2}}$
= $ \sqrt{(2)^{2}+(2)^{2}}$
=$ \sqrt{4+4}$
= $ \sqrt{8}$
=$2\sqrt{2}$
BC=$\sqrt{(-1-1)^{2}+(2-0)^{2}}$
=$ \sqrt{(-2)^{2}+(2)^{2}}$
=$\sqrt{4+4}$
= $ \sqrt{8}$
=$2\sqrt{2}$
CD= $\sqrt{(-3+1)^{2}+(0-2)^{2}}$
=$\sqrt{(-2)^{2}+(-2)^{2}}$
=$\sqrt{4+4}$
=$\sqrt{8}$
=$2\sqrt{2}$
AD=$ \sqrt{(-3+1)^{2}+(0-2)^{2}}$
=$\sqrt{(-2)^{2}+(-2)^{2}}$
=$\sqrt{4+4}$
= $\sqrt{8}$
=$2\sqrt{2}$
Diagonal AC=$\sqrt{(-1+1)^{2}+(2+2)^{2}}$
=$\sqrt{(0)^{2}+(4)^{2}}$
=$\sqrt{16}$
=4
Diagonal BD=$\sqrt{(-3-1)^{2}+(0-0)^{2}}$
=$\sqrt{(-4)^{2}+(0)^{2}}$
=$\sqrt{16}$
=4
Side length = AB = BC = CD = DA =$2\sqrt{2}$
Diagonal Measure = AC = BD = 4
Therefore, the given points are the vertices of a square.
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)
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=$\sqrt{(-3-3)^{2}+(1-5)^{2}}$
=$\sqrt{(-6)^{2}+(-4)^{2}}$
= $\sqrt{36+16}$
=$\sqrt{52}$
= $2\sqrt{13}$
BC= $\sqrt{(0-3)^{2}+(3-1)^{2}}$
=$\sqrt{(-3)^{2}+(2)^{2}}$
= $\sqrt{9+4}$
= $\sqrt{13}$
CD= $ \sqrt{(-1-0)^{2}+(-4-3)^{2}}$
= $\sqrt{(1)^{2}+(7)^{2}}$
=$ \sqrt{1+49}$
=$ \sqrt{50}$
= $5\sqrt{2}$
AD= $\sqrt{(-1+3)^{2}+(-4-5)^{2}}$
= $\sqrt{(2)^{2}+(-9)^{2}}$
=$ \sqrt{4+81}$
= $\sqrt{85}$
Its also seen 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 which has 4 sides.
Therefore, the given points cannot form a general quadrilateral.
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
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=$\sqrt{(7-4)^{2}+(6-5)^{2}}$
=$\sqrt{(3)^{2}+(-1)^{2}}$
=$ \sqrt{9+1}$
= $ \sqrt{10}$
BC=$\sqrt{(4-7)^{2}+(3-6)^{2}}$
= $\sqrt{(-3)^{2}+(-3)^{2}}$
= $\sqrt{9+8}$
= $\sqrt{18}$
CD=$\sqrt{(1-4)^{2}+(2-3)^{2}}$
= $\sqrt{(-3)^{2}+(-1)^{2}}$
= $\sqrt{9+1}$
= $ \sqrt{10}$
AD= $\sqrt{(1-4)^{2}+(2-5)^{2}}$
=$ \sqrt{(-3)^{2}+(-3)^{2}}$
=$\sqrt{9+9}$
= $ \sqrt{18}$
Diagonal AC= $\sqrt{(4-4)^{2}+(3-5)^{2}}$
= $\sqrt{(0)^{2}+(-2)^{2}}$
=$ \sqrt{0+4}$
=2
Diagonal BD=$\sqrt{(1-7)^{2}+(2-6)^{2}}$
=$ \sqrt{(-6)^{2}+(-4)^{2}}$
=$ \sqrt{36+16}$
=$\sqrt{52}$
=$13\sqrt{2}$
Opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths.
Therefore, the given points are
the vertices of a parallelogram.
We have to find point on x axis.So,its y coordinate will be 0.
Let point on x-axis be (x,0)
Distance between (x,0) and (2,-5)
= $\sqrt{(2-x)^{2}+(-5-0)^{2}}$
= $\sqrt{(2-x)^{2}+(-5)^{2}}$
= $\sqrt{(2-x)^{2}+(25)}$
Distance between (x,0) and (-2,9)
= $ \sqrt{(-2-x)^{2}+(9-0)^{2}}$
= $\sqrt{(-2-x)^{2}+(9)^{2}}$
= $\sqrt{(-2-x)^{2}+(81)}$
By given condition these distances are equal in measure.
$\sqrt{(2-x)^{2}+(25)}$= $\sqrt{(-2-x)^{2}+(81)}$
Simplify the above equation, Remove square root by taking square both the sides, we get $(2-x)^{2}+(25)=(-2-x)^{2}+(81)$ $(2-x)^{2}+(25)=[-(2+x)]^{2}+(81)$ $(2-x)^{2}+(25)=(2+x)^{2}+(81)$$(x)^{2}+4-4x+25$=$ (x)^{2}+4+4x+81$
8x=25-81
8x=25-81
8x=-56
x=-7
Therefore the point is(-7,0).
Given that distance between (2,-3) and (10,y) is 10
Therefore using distance formula
PQ=$\sqrt{(10-2)^{2}+(y+3)^{2}}$=10
$\sqrt{(8)^{2}+(y+3)^{2}}$=10
Simplify the above equation and find the value of y. Squaring both sides,$64+(y+3)^{2}$=100
$(y+3)^{2}$=36
y+3=$\pm$ 6
y+3=6 or y+3=-6
Therefore y=3 or -9
Given: Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), which means PQ = QR
Step 1: Find the distance between PQ and QR using distance formula,
PQ=$\sqrt{(5-0)^{2}+(-3-1)^{2}} $
=$\sqrt{(5)^{2}+(-4)^{2}} $
= $\sqrt{25+16} $
= 41
QR = $ \sqrt{(0-x)^{2}+(1-6)^{2}}$
= $ \sqrt{(-x)^{2}+(-5)^{2}}$
= $ \sqrt{(x)^{2}+25}$
=$x^{2}+25$
Step 2: Use PQ=QR
$\sqrt{41}$=$\sqrt{x^{2}+25}$
Squaring both the sides, to omit square root
41 = x2 + 25
$x^{2}$ = 16
x = ± 4
x = 4 or x = -4
Coordinates of Point R will bet R is (4,6) or (-4,6)
When point R is (4,6)
PR= $\sqrt{(5-4)^{2}+(-3-6)^{2}}$
=$ \sqrt{(1)^{2}+(-9)^{2}} $
= $ \sqrt{1+81} $
=$ \sqrt{82}$
QR=$ \sqrt{(0-4)^{2}+(1-6)^{2}} $
= $\sqrt{(-4)^{2}+(-5)^{2}} $
=$ \sqrt{16+25} $
= $\sqrt{41} $
When point R is (-4,6)
PR=$\sqrt{(5-(-4))^{2}+(-3-6)^{2}}$
=$ \sqrt{(9)^{2}+(-9)^{2}}$
= $\sqrt{81+81}$
=9$ \sqrt{2} $
QR= $\sqrt{(0-(-4))^{2}+(1-6)^{2}}$
=$ \sqrt{(4)^{2}+(-5)^{2}} $
=$\sqrt{16+25} $
= $\sqrt{41} $
Point (x,y) is equidistant from (3,6) and (-3,4)
Therefore $\sqrt{(x-3)^{2}+(y-6)^{2}}$ = $ \sqrt{(x-(-3))^{2}+(y-4)^{2}} $
$ \sqrt{(x-3)^{2}+(y-6)^{2}}$ = $\sqrt{(x+3)^{2}+(y-4)^{2}} $
$ (x-3)^{2}+(y-6)^{2}$ =$(x+3)^{2}+(y-4)^{2}$
$x^{2}+9-6x+y^{2}+36-12y$=$x^{2}+9+6x+y^{2}+16-8y$
36-16=6x+6x+12y-8y
20=12x+4y
3x+y=5
3x+y-5=0