Solution

(i) Distance between the two points is given by

$ \sqrt{(x_1-x_2)^{2}+(y_1-y_2)^{2}} $

Therefore, distance between $(2,3)$ and $(4,1)$ is given by

$ \begin{aligned} l=\sqrt{(2-4)^{2}+(3-1)^{2}} & =\sqrt{(-2)^{2}+(2)^{2}} \\ & =\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} \end{aligned} $

(ii) Distance between $(-5,7)$ and $(-1,3)$ is given by

$ \begin{aligned} l & =\sqrt{(-5-(-1))^{2}+(7-3)^{2}}=\sqrt{(-4)^{2}+(4)^{2}} \\ & =\sqrt{16+16}=\sqrt{32}=4 \sqrt{2} \end{aligned} $

(iii) Distance between $(a, b)$ and $(-a,-b)$ is given by

$ \begin{aligned} & l=\sqrt{(a-(-a))^{2}+(b-(-b))^{2}} \\ & =\sqrt{(2 a)^{2}+(2 b)^{2}}=\sqrt{4 a^{2}+4 b^{2}}=2 \sqrt{a^{2}+b^{2}} \end{aligned} $

Solution

Distance between points $(0,0)$ and $(36,15)$

$ \begin{aligned} & =\sqrt{(36-0)^{2}+(15-0)^{2}}=\sqrt{36^{2}+15^{2}} \\ & =\sqrt{1296+225}=\sqrt{1521}=39 \end{aligned} $

Yes, we can find the distance between the given towns A and B.

Assume town $A$ at origin point $(0,0)$.

Therefore, town B will be at point $(36,15)$ with respect to town A.

And hence, as calculated above, the distance between town $A$ and $B$ will be $39 km$.

Solution

Let the points $(1,5),(2,3)$, and $(-2,-11)$ be representing the vertices $A, B$, and $C$ of the given triangle respectively.

Let $A=(1,5), B=(2,3), C=(-2,-11)$

$\therefore AB=\sqrt{(1-2)^{2}+(5-3)^{2}}=\sqrt{5}$

$BC=\sqrt{(2-(-2))^{2}+(3-(-11))^{2}}=\sqrt{4^{2}+14^{2}}=\sqrt{16+196}=\sqrt{212}$

$CA=\sqrt{(1-(-2))^{2}+(5-(-11))^{2}}=\sqrt{3^{2}+16^{2}}=\sqrt{9+256}=\sqrt{265}$

Since $AB+BC \neq CA$,

Therefore, the points $(1,5),(2,3)$, and ( - 2, - 11) are not collinear.

Solution

Let the points $(5,-2),(6,4)$, and $(7,-2)$ are representing the vertices $A, B$, and $C$ of the given triangle respectively.

$ \begin{aligned} & AB=\sqrt{(5-6)^{2}+(-2-4)^{2}}=\sqrt{(-1)^{2}+(-6)^{2}}=\sqrt{1+36}=\sqrt{37} \\ & BC=\sqrt{(6-7)^{2}+(4-(-2))^{2}}=\sqrt{(-1)^{2}+(6)^{2}}=\sqrt{1+36}=\sqrt{37} \\ & CA=\sqrt{(5-7)^{2}+(-2-(-2))^{2}}=\sqrt{(-2)^{2}+0^{2}}=2 \end{aligned} $

Therefore, $AB=BC$

As two sides are equal in length, therefore, $ABC$ is an isosceles triangle.

Solution

It can be observed that $A(3,4), B(6,7), C(9,4)$, and $D(6,1)$ are the positions of these 4 friends.

$AB=\sqrt{(3-6)^{2}+(4-7)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$

$BC=\sqrt{(6-9)^{2}+(7-4)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$

$CB=\sqrt{(9-6)^{2}+(4-1)^{2}}=\sqrt{(3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$

$AD=\sqrt{(3-6)^{2}+(4-1)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$

Diagonal $AC=\sqrt{(3-9)^{2}+(4-4)^{2}}=\sqrt{(-6)^{2}+0^{2}}=6$

Diagonal $BD=\sqrt{(6-6)^{2}+(7-1)^{2}}=\sqrt{0^{2}+(6)^{2}}=6$

It can be observed that all sides of this quadrilateral $A B C D$ are of the same length and also the diagonals are of the same length.

Therefore, $A B C D$ is a square and hence, Champa was correct

Solution

(i) 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.

$\therefore AB=\sqrt{(-1-1)^{2}+(-2-0)^{2}}=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$

$BC=\sqrt{(1-(-1))^{2}+(0-2)^{2}}=\sqrt{(2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$

$CD=\sqrt{(-1-(-3))^{2}+(2-0)^{2}}=\sqrt{(2)^{2}+(2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$

$AD=\sqrt{(-1-(-3))^{2}+(-2-0)^{2}}=\sqrt{(2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$

Diagonal $AC=\sqrt{(-1-(-1))^{2}+(-2-2)^{2}}=\sqrt{0^{2}+(-4)^{2}}=\sqrt{16}=4$

Diagonal $BD=\sqrt{(1-(-3))^{2}+(0-0)^{2}}=\sqrt{(4)^{2}+0^{2}}=\sqrt{16}=4$

It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.

(ii)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.

$ \begin{aligned} & AB=\sqrt{(-3-3)^{2}+(5-1)^{2}}=\sqrt{(-6)^{2}+(4)^{2}}=\sqrt{36+16}=\sqrt{52}=2 \sqrt{13} \\ & BC=\sqrt{(3-0)^{2}+(1-3)^{2}}=\sqrt{(3)^{2}+(-2)^{2}}=\sqrt{9+4}=\sqrt{13} \\ & CD=\sqrt{(0-(-1))^{2}+(3-(-4))^{2}}=\sqrt{(1)^{2}+(7)^{2}}=\sqrt{1+49}=\sqrt{50}=5 \sqrt{2} \\ & AD=\sqrt{(-3-(-1))^{2}+(5-(-4))^{2}}=\sqrt{(-2)^{2}+(9)^{2}}=\sqrt{4+81}=\sqrt{85} \end{aligned} $

It can be observed that all sides of this quadrilateral are of different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc.

(iii)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.

$ \begin{aligned} & AB=\sqrt{(4-7)^{2}+(5-6)^{2}}=\sqrt{(-3)^{2}+(-1)^{2}}=\sqrt{9+1}=\sqrt{10} \\ & BC=\sqrt{(7-4)^{2}+(6-3)^{2}}=\sqrt{(3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18} \\ & CD=\sqrt{(4-1)^{2}+(3-2)^{2}}=\sqrt{(3)^{2}+(1)^{2}}=\sqrt{9+1}=\sqrt{10} \\ & AD=\sqrt{(4-1)^{2}+(5-2)^{2}}=\sqrt{(3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18} \end{aligned} $

Diagonal $AC=\sqrt{(4-4)^{2}+(5-3)^{2}}=\sqrt{(0)^{2}+(2)^{2}}=\sqrt{0+4}=2$

Diagonal $CD=\sqrt{(7-1)^{2}+(6-2)^{2}}=\sqrt{(6)^{2}+(4)^{2}}=\sqrt{36+16}=\sqrt{52}=13 \sqrt{2}$

It can be observed that 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.

Solution

We have to find a point on $x$-axis. Therefore, its $y$-coordinate will be 0 .

Let the point on $x$-axis be $(x, 0)$.

Distance between $(x, 0)$ and $(2,-5)=\sqrt{(x-2)^{2}+(0-(-5))^{2}}=\sqrt{(x-2)^{2}+(5)^{2}}$

Distance between $(x, 0)$ and $(-2,9)=\sqrt{(x-(-2))^{2}+(0-(-9))^{2}}=\sqrt{(x+2)^{2}+(9)^{2}}$

By the given condition, these distances are equal in measure. $\sqrt{(x-2)^{2}+(5)^{2}}=\sqrt{(x+2)^{2}+(9)^{2}}$

$(x-2)^{2}+25=(x+2)^{2}+81$

$x^{2}+4-4 x+25=x^{2}+4+4 x+81$

$8 x=25-81$

$8 x=-56$

$x=-7$

Therefore, the point is $(-7,0)$.

Solution

It is given that the distance between $(2,-3)$ and $(10, y)$ is 10 .

Therefore, $\sqrt{(2-10)^{2}+(-3-y)^{2}}=10$

$\sqrt{(-8)^{2}+(3+y)^{2}}=10$

$64+(y+3)^{2}=100$

$(y+3)^{2}=36$

$y+3= \pm 6$

$y+3=6$ or $y+3=-6$

Therefore, $y=3$ or -9

Solution

$PQ=QR$

$\sqrt{(5-0)^{2}+(-3-1)^{2}}=\sqrt{(0-x)^{2}+(1-6)^{2}}$

$\sqrt{(5)^{2}+(-4)^{2}}=\sqrt{(-x)^{2}+(-5)^{2}}$

$\sqrt{25+16}=\sqrt{x^{2}+25}$

$41=x^{2}+25$

$16=x^{2}$

$x= \pm 4$

Therefore, point $R$ is $(4,6)$ or $(-4,6)$.

When point $R$ is $(4,6)$,

$ \begin{aligned} & 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} \end{aligned} $

When point $R$ is $(-4,6)$,

$ \begin{aligned} & 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} \end{aligned} $

Solution

Point $(x, y)$ is equidistant from $(3,6)$ and $(-3,4)$.

$ \begin{aligned} & \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}+9-6 x+y^{2}+36-12 y=x^{2}+9+6 x+y^{2}+16-8 y \\ & 36-16=6 x+6 x+12 y-8 y \\ & 20=12 x+4 y \\ & 3 x+y=5 \\ & 3 x+y-5=0 \end{aligned} $