Distance Formula Questions in English

(N/A) Let $A(0, -3), B(1, -1),$ and $C(2, 1)$ be the given points.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(1 - 0)^2 + (-1 - (-3))^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$
$BC = \sqrt{(2 - 1)^2 + (1 - (-1))^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$
$AC = \sqrt{(2 - 0)^2 + (1 - (-3))^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$
Since $AB + BC = \sqrt{5} + \sqrt{5} = 2\sqrt{5} = AC$,the sum of the lengths of two segments is equal to the length of the third segment.
Therefore,the points $A, B,$ and $C$ are collinear.

(A) Given the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Here,$A = (2, 4)$,$B = (-3, b)$,and $AB = \sqrt{26}$.
Squaring both sides,we get $AB^2 = 26$.
Substituting the coordinates: $(2 - (-3))^2 + (4 - b)^2 = 26$.
$(2 + 3)^2 + (4 - b)^2 = 26$.
$5^2 + (16 - 8b + b^2) = 26$.
$25 + 16 - 8b + b^2 = 26$.
$b^2 - 8b + 41 = 26$.
$b^2 - 8b + 15 = 0$.
Factoring the quadratic equation: $(b - 5)(b - 3) = 0$.
Therefore,$b = 5$ or $b = 3$.

(A-D) Let the vertices be $A(1, 7), B(2, 4)$,and $C(k, 5)$.
Calculate the squared distances between the points:
$AB^2 = (1-2)^2 + (7-4)^2 = (-1)^2 + 3^2 = 1 + 9 = 10$
$BC^2 = (2-k)^2 + (4-5)^2 = 4 - 4k + k^2 + 1 = k^2 - 4k + 5$
$AC^2 = (1-k)^2 + (7-5)^2 = 1 - 2k + k^2 + 4 = k^2 - 2k + 5$
For a right triangle,one of the angles must be $90^{\circ}$:
Case $1$: $\angle A = 90^{\circ}$. Then $BC^2 = AB^2 + AC^2$.
$k^2 - 4k + 5 = 10 + k^2 - 2k + 5$
$-2k = 10 \implies k = -5$
Case $2$: $\angle B = 90^{\circ}$. Then $AC^2 = AB^2 + BC^2$.
$k^2 - 2k + 5 = 10 + k^2 - 4k + 5$
$2k = 10 \implies k = 5$
Case $3$: $\angle C = 90^{\circ}$. Then $AB^2 = BC^2 + AC^2$.
$10 = (k^2 - 4k + 5) + (k^2 - 2k + 5)$
$10 = 2k^2 - 6k + 10$
$2k^2 - 6k = 0 \implies 2k(k - 3) = 0$
$k = 0$ or $k = 3$
Thus,the possible values of $k$ are $-5, 5, 0, 3$.

(N/A) Let the vertices be $A(0,6), B(-5,3)$,and $C(3,1)$.
Using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$:
$AB^2 = (-5 - 0)^2 + (3 - 6)^2 = (-5)^2 + (-3)^2 = 25 + 9 = 34$
$BC^2 = (3 - (-5))^2 + (1 - 3)^2 = (8)^2 + (-2)^2 = 64 + 4 = 68$
$AC^2 = (3 - 0)^2 + (1 - 6)^2 = (3)^2 + (-5)^2 = 9 + 25 = 34$
Since $AB^2 + AC^2 = 34 + 34 = 68 = BC^2$,the triangle satisfies the Pythagorean theorem,so it is a right-angled triangle with $\angle BAC = 90^\circ$.
Also,since $AB^2 = AC^2 = 34$,we have $AB = AC$.
Therefore,the triangle is an isosceles right triangle.

(A) Let the vertices of the equilateral triangle be $A(0,0)$,$B(3, \sqrt{3})$,and $C(x, y)$.
Since it is an equilateral triangle,$AB = BC = AC$,which implies $AB^2 = BC^2 = AC^2$.
First,calculate $AB^2 = (3-0)^2 + (\sqrt{3}-0)^2 = 9 + 3 = 12$.
Since $AC^2 = AB^2$,we have $x^2 + y^2 = 12$ (Equation $1$).
Since $BC^2 = AB^2$,we have $(x-3)^2 + (y-\sqrt{3})^2 = 12$.
Expanding this: $x^2 - 6x + 9 + y^2 - 2\sqrt{3}y + 3 = 12$.
Substituting $x^2 + y^2 = 12$ into this equation: $12 - 6x - 2\sqrt{3}y + 12 = 12$,which simplifies to $6x + 2\sqrt{3}y = 12$,or $y = \sqrt{3}(2-x)$ (Equation $2$).
Substituting Equation $2$ into Equation $1$: $x^2 + [\sqrt{3}(2-x)]^2 = 12$.
$x^2 + 3(4 - 4x + x^2) = 12$.
$x^2 + 12 - 12x + 3x^2 = 12$.
$4x^2 - 12x = 0$,so $4x(x-3) = 0$.
This gives $x = 0$ or $x = 3$.
If $x = 0$,$y = \sqrt{3}(2-0) = 2\sqrt{3}$.
If $x = 3$,$y = \sqrt{3}(2-3) = -\sqrt{3}$.
Thus,the third vertex is $(0, 2\sqrt{3})$ or $(3, -\sqrt{3})$.

(N/A) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
First,calculate $SP$:
$SP = \sqrt{(a - at^2)^2 + (0 - 2at)^2} = \sqrt{a^2(1-t^2)^2 + 4a^2t^2} = |a|\sqrt{1 - 2t^2 + t^4 + 4t^2} = |a|\sqrt{(1+t^2)^2} = |a|(1+t^2)$.
Next,calculate $SQ$:
$SQ = \sqrt{(a - \frac{a}{t^2})^2 + (0 - (-\frac{2a}{t}))^2} = \sqrt{a^2(1 - \frac{1}{t^2})^2 + \frac{4a^2}{t^2}} = |a|\sqrt{1 - \frac{2}{t^2} + \frac{1}{t^4} + \frac{4}{t^2}} = |a|\sqrt{1 + \frac{2}{t^2} + \frac{1}{t^4}} = |a|\sqrt{(1 + \frac{1}{t^2})^2} = |a|(1 + \frac{1}{t^2}) = |a|(\frac{t^2+1}{t^2})$.
Now,evaluate $\frac{1}{SP} + \frac{1}{SQ}$:
$\frac{1}{SP} + \frac{1}{SQ} = \frac{1}{|a|(1+t^2)} + \frac{1}{|a|(\frac{t^2+1}{t^2})} = \frac{1}{|a|(1+t^2)} + \frac{t^2}{|a|(1+t^2)} = \frac{1+t^2}{|a|(1+t^2)} = \frac{1}{|a|}$.
Since the result $\frac{1}{|a|}$ does not contain $t$,the expression $\frac{1}{SP} + \frac{1}{SQ}$ is independent of $t$.

(N/A) To show that points $A(-1, 4)$,$B(2, 3)$,and $C(5, 2)$ are collinear,we must show that the sum of the lengths of two segments equals the length of the third segment.
$1$. Calculate the distance $AB$ using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(2 - (-1))^2 + (3 - 4)^2} = \sqrt{(3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.
$2$. Calculate the distance $BC$:
$BC = \sqrt{(5 - 2)^2 + (2 - 3)^2} = \sqrt{(3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.
$3$. Calculate the distance $AC$:
$AC = \sqrt{(5 - (-1))^2 + (2 - 4)^2} = \sqrt{(6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$.
$4$. Since $AB + BC = \sqrt{10} + \sqrt{10} = 2\sqrt{10} = AC$,the points $A$,$B$,and $C$ lie on the same line and are therefore collinear.

(N/A) To prove that point $P(7, 5)$ is equidistant from $A(2, 4)$ and $B(6, 10)$,we need to show that the distance $PA$ is equal to the distance $PB$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$1$. Distance $PA = \sqrt{(7 - 2)^2 + (5 - 4)^2} = \sqrt{5^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26}$.
$2$. Distance $PB = \sqrt{(7 - 6)^2 + (5 - 10)^2} = \sqrt{1^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26}$.
Since $PA = PB = \sqrt{26}$,the point $P(7, 5)$ is equidistant from points $A(2, 4)$ and $B(6, 10)$.

(A) Let the point on the $X$-axis be $P(x, 0)$.
The distance between $P(x, 0)$ and $A(11, 12)$ is given as $13$.
Using the distance formula: $\sqrt{(x - 11)^2 + (0 - 12)^2} = 13$.
Squaring both sides: $(x - 11)^2 + (-12)^2 = 13^2$.
$(x - 11)^2 + 144 = 169$.
$(x - 11)^2 = 169 - 144 = 25$.
Taking the square root: $x - 11 = \pm 5$.
Case $1$: $x - 11 = 5 \implies x = 16$.
Case $2$: $x - 11 = -5 \implies x = 6$.
Therefore,the points are $(16, 0)$ and $(6, 0)$.

(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $A(a, 2)$ and $B(3, -5)$ with distance $d = \sqrt{53}$.
Substituting the values into the formula:
$\sqrt{(3 - a)^2 + (-5 - 2)^2} = \sqrt{53}$
Squaring both sides:
$(3 - a)^2 + (-7)^2 = 53$
$(3 - a)^2 + 49 = 53$
$(3 - a)^2 = 53 - 49$
$(3 - a)^2 = 4$
Taking the square root on both sides:
$3 - a = 2$ or $3 - a = -2$
If $3 - a = 2$,then $a = 1$.
If $3 - a = -2$,then $a = 5$.
Therefore,the possible values of $a$ are $1$ and $5$.

(N/A) Let the given points be $A(1, 1)$,$B(4, 4)$,and $C(6, 2)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$1$. Length of $AB = \sqrt{(4 - 1)^2 + (4 - 1)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18}$.
$2$. Length of $BC = \sqrt{(6 - 4)^2 + (2 - 4)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}$.
$3$. Length of $AC = \sqrt{(6 - 1)^2 + (2 - 1)^2} = \sqrt{5^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26}$.
Now,check for the Pythagorean theorem $(AB^2 + BC^2 = AC^2)$:
$AB^2 = 18$,$BC^2 = 8$,and $AC^2 = 26$.
Since $18 + 8 = 26$,we have $AB^2 + BC^2 = AC^2$.
Therefore,the triangle is a right-angled triangle with the right angle at vertex $B$.

(N/A) To prove that the points $A(1, 7)$,$B(2, 4)$,and $C(5, 5)$ form an isosceles right triangle,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of $AB$: $AB = \sqrt{(2 - 1)^2 + (4 - 7)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$.
$2$. Length of $BC$: $BC = \sqrt{(5 - 2)^2 + (5 - 4)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$.
$3$. Length of $AC$: $AC = \sqrt{(5 - 1)^2 + (5 - 7)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20}$.
Since $AB = BC = \sqrt{10}$,the triangle is isosceles.
Now,check for the right-angled condition using the Pythagorean theorem $(AB^2 + BC^2 = AC^2)$:
$AB^2 + BC^2 = 10 + 10 = 20$.
$AC^2 = 20$.
Since $AB^2 + BC^2 = AC^2$,the triangle is a right-angled triangle.
Thus,$A, B,$ and $C$ are the vertices of an isosceles right triangle.

(N/A) To prove that the points $A(0,0), B(7,0), C(7,5)$,and $D(0,5)$ form a rectangle,we must show that opposite sides are equal and the diagonals are equal.
$1$. Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(7-0)^2 + (0-0)^2} = \sqrt{49} = 7$
$BC = \sqrt{(7-7)^2 + (5-0)^2} = \sqrt{25} = 5$
$CD = \sqrt{(0-7)^2 + (5-5)^2} = \sqrt{49} = 7$
$DA = \sqrt{(0-0)^2 + (0-5)^2} = \sqrt{25} = 5$
Since $AB = CD = 7$ and $BC = DA = 5$,opposite sides are equal.
$2$. Calculate the lengths of the diagonals:
$AC = \sqrt{(7-0)^2 + (5-0)^2} = \sqrt{49 + 25} = \sqrt{74}$
$BD = \sqrt{(0-7)^2 + (5-0)^2} = \sqrt{49 + 25} = \sqrt{74}$
Since the opposite sides are equal and the diagonals are equal,the quadrilateral $ABCD$ is a rectangle.

(N/A) Let the vertices be $A(1, -2)$,$B(2, 3)$,$C(-3, 2)$,and $D(-4, -3)$.
To prove that $ABCD$ is a rhombus,we must show that all four sides are equal in length,but the diagonals are not equal.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(2 - 1)^2 + (3 - (-2))^2} = \sqrt{1^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}$
$BC = \sqrt{(-3 - 2)^2 + (2 - 3)^2} = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26}$
$CD = \sqrt{(-4 - (-3))^2 + (-3 - 2)^2} = \sqrt{(-1)^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26}$
$DA = \sqrt{(1 - (-4))^2 + (-2 - (-3))^2} = \sqrt{5^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26}$
Since $AB = BC = CD = DA = \sqrt{26}$,all sides are equal.
Now,calculate the lengths of the diagonals $AC$ and $BD$:
$AC = \sqrt{(-3 - 1)^2 + (2 - (-2))^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
$BD = \sqrt{(-4 - 2)^2 + (-3 - 3)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$
Since all sides are equal $(AB=BC=CD=DA)$ and the diagonals are not equal $(AC \neq BD)$,the quadrilateral $ABCD$ is a rhombus.

(A) Let the given points be $A(2, 2)$,$B(-2, 2)$,$C(-2, -2)$,and $D(2, -2)$.
To prove that these points form a square,we must show that all four sides are equal and both diagonals are equal.
$1$. Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(-2 - 2)^2 + (2 - 2)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4$ units.
$BC = \sqrt{(-2 - (-2))^2 + (-2 - 2)^2} = \sqrt{0^2 + (-4)^2} = \sqrt{16} = 4$ units.
$CD = \sqrt{(2 - (-2))^2 + (-2 - (-2))^2} = \sqrt{4^2 + 0^2} = \sqrt{16} = 4$ units.
$DA = \sqrt{(2 - 2)^2 + (2 - (-2))^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4$ units.
Since $AB = BC = CD = DA = 4$,all sides are equal.
$2$. Calculate the lengths of the diagonals:
$AC = \sqrt{(-2 - 2)^2 + (-2 - 2)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$ units.
$BD = \sqrt{(2 - (-2))^2 + (-2 - 2)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$ units.
Since the sides are equal and the diagonals are equal,the points $A, B, C, D$ form a square.

(A) Let the vertices be $A(1, -3/2)$,$B(-3, -7/2)$,and $C(-4, -3/2)$.
To check if it is a right-angled triangle,we calculate the squares of the lengths of the sides using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.
$AB^2 = (-3 - 1)^2 + (-7/2 - (-3/2))^2 = (-4)^2 + (-4/2)^2 = 16 + (-2)^2 = 16 + 4 = 20$.
$BC^2 = (-4 - (-3))^2 + (-3/2 - (-7/2))^2 = (-1)^2 + (4/2)^2 = 1 + 2^2 = 1 + 4 = 5$.
$AC^2 = (-4 - 1)^2 + (-3/2 - (-3/2))^2 = (-5)^2 + (0)^2 = 25 + 0 = 25$.
Since $AB^2 + BC^2 = 20 + 5 = 25$,which is equal to $AC^2$,the triangle satisfies the Pythagorean theorem.
Therefore,the given points form a right-angled triangle.

(N/A) Given that point $P(x, y)$ is equidistant from points $A(a+b, b-a)$ and $B(a-b, a+b).$
By the distance formula,$PA = PB.$
Squaring both sides,$PA^2 = PB^2.$
Using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2,$
$(x - (a+b))^2 + (y - (b-a))^2 = (x - (a-b))^2 + (y - (a+b))^2.$
Expanding both sides:
$x^2 - 2x(a+b) + (a+b)^2 + y^2 - 2y(b-a) + (b-a)^2 = x^2 - 2x(a-b) + (a-b)^2 + y^2 - 2y(a+b) + (a+b)^2.$
Canceling $x^2, y^2, (a+b)^2$ from both sides:
$-2x(a+b) - 2y(b-a) + (b-a)^2 = -2x(a-b) - 2y(a+b) + (a-b)^2.$
Since $(b-a)^2 = (a-b)^2,$ these terms also cancel out:
$-2ax - 2bx - 2by + 2ay = -2ax + 2bx - 2ay - 2by.$
Canceling $-2ax$ and $-2by$ from both sides:
$-2bx + 2ay = 2bx - 2ay.$
Rearranging the terms:
$2ay + 2ay = 2bx + 2bx.$
$4ay = 4bx.$
Dividing by $4,$ we get $bx = ay.$

(C) Let the point on the $Y$-axis be $P(0, y)$.
Given points are $A(-5, -2)$ and $B(3, 2)$.
Since $P$ is equidistant from $A$ and $B$,we have $PA = PB$.
Using the distance formula,$PA^2 = PB^2$.
$(0 - (-5))^2 + (y - (-2))^2 = (0 - 3)^2 + (y - 2)^2$
$5^2 + (y + 2)^2 = (-3)^2 + (y - 2)^2$
$25 + y^2 + 4y + 4 = 9 + y^2 - 4y + 4$
$29 + 4y = 13 - 4y$
$8y = 13 - 29$
$8y = -16$
$y = -2$
Therefore,the required point is $(0, -2)$.

(N/A) Let the vertices be $A(2a, 4a)$, $B(2a, 6a)$, and $C(2a + \sqrt{3}a, 5a)$.
To prove that the triangle is equilateral, we must show that the lengths of all three sides are equal.
The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of side $AB$:
$AB = \sqrt{(2a - 2a)^2 + (6a - 4a)^2} = \sqrt{0^2 + (2a)^2} = \sqrt{4a^2} = 2a$.
$2$. Length of side $BC$:
$BC = \sqrt{(2a + \sqrt{3}a - 2a)^2 + (5a - 6a)^2} = \sqrt{(\sqrt{3}a)^2 + (-a)^2} = \sqrt{3a^2 + a^2} = \sqrt{4a^2} = 2a$.
$3$. Length of side $AC$:
$AC = \sqrt{(2a + \sqrt{3}a - 2a)^2 + (5a - 4a)^2} = \sqrt{(\sqrt{3}a)^2 + (a)^2} = \sqrt{3a^2 + a^2} = \sqrt{4a^2} = 2a$.
Since $AB = BC = AC = 2a$, the triangle $ABC$ is an equilateral triangle.

(A) Let the vertices of the triangle be $A(-1, 3)$,$B(1, -1)$,and $C(5, 1)$.
The median through vertex $A$ connects $A$ to the midpoint of the opposite side $BC$.
Let $M$ be the midpoint of $BC$. The coordinates of $M$ are given by the midpoint formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{1 + 5}{2}, \frac{-1 + 1}{2} \right) = \left( \frac{6}{2}, \frac{0}{2} \right) = (3, 0)$.
The length of the median $AM$ is the distance between $A(-1, 3)$ and $M(3, 0)$.
Using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$AM = \sqrt{(3 - (-1))^2 + (0 - 3)^2} = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Thus,the length of the median is $5$.

(P(2, -1)) Let $P(x, y)$ be the circumcentre of $\Delta ABC$.
Since $P$ is the circumcentre,the distance from $P$ to each vertex is equal,i.e.,$PA = PB = PC$,which implies $PA^2 = PB^2 = PC^2$.
$1$. $PA^2 = (x + 3)^2 + (y + 1)^2 = x^2 + 6x + 9 + y^2 + 2y + 1 = x^2 + y^2 + 6x + 2y + 10$
$2$. $PB^2 = (x + 1)^2 + (y - 3)^2 = x^2 + 2x + 1 + y^2 - 6y + 9 = x^2 + y^2 + 2x - 6y + 10$
$3$. $PC^2 = (x - 6)^2 + (y - 2)^2 = x^2 - 12x + 36 + y^2 - 4y + 4 = x^2 + y^2 - 12x - 4y + 40$
Equating $PA^2 = PB^2$:
$x^2 + y^2 + 6x + 2y + 10 = x^2 + y^2 + 2x - 6y + 10$
$4x + 8y = 0 \implies x = -2y$ ... $(1)$
Equating $PB^2 = PC^2$:
$x^2 + y^2 + 2x - 6y + 10 = x^2 + y^2 - 12x - 4y + 40$
$14x - 2y = 30 \implies 7x - y = 15$ ... $(2)$
Substitute $(1)$ into $(2)$:
$7(-2y) - y = 15$
$-14y - y = 15 \implies -15y = 15 \implies y = -1$
Using $y = -1$ in $(1)$:
$x = -2(-1) = 2$
Thus,the circumcentre is $P(2, -1)$.

(C) Let the circumcentre be $O(x, y)$. The circumcentre is equidistant from the vertices $A, B$,and $C$. Therefore,$OA^2 = OB^2 = OC^2$.
Given $A(-1, 1)$,$B(0, -4)$,and $C(-1, -5)$.
$OA^2 = (x + 1)^2 + (y - 1)^2 = x^2 + 2x + 1 + y^2 - 2y + 1 = x^2 + y^2 + 2x - 2y + 2$.
$OB^2 = (x - 0)^2 + (y + 4)^2 = x^2 + y^2 + 8y + 16$.
$OC^2 = (x + 1)^2 + (y + 5)^2 = x^2 + 2x + 1 + y^2 + 10y + 25 = x^2 + y^2 + 2x + 10y + 26$.
Equating $OA^2 = OC^2$:
$x^2 + y^2 + 2x - 2y + 2 = x^2 + y^2 + 2x + 10y + 26$
$-2y + 2 = 10y + 26$
$-12y = 24 \implies y = -2$.
Equating $OA^2 = OB^2$:
$x^2 + y^2 + 2x - 2y + 2 = x^2 + y^2 + 8y + 16$
$2x - 2y + 2 = 8y + 16$
$2x - 10y = 14$.
Substituting $y = -2$:
$2x - 10(-2) = 14$
$2x + 20 = 14$
$2x = -6 \implies x = -3$.
Thus,the circumcentre is $(-3, -2)$.

(B) Let the vertices of the triangle be $A(a, -1)$,$B(6, -9)$,and $C(10, b)$. Let the circumcentre be $O(6, -5)$.
Since the circumcentre is equidistant from all vertices,$OA = OB = OC$.
First,calculate $OB^2 = (6 - 6)^2 + (-5 - (-9))^2 = 0^2 + 4^2 = 16$.
Since $OA^2 = OB^2$,we have $(a - 6)^2 + (-1 - (-5))^2 = 16$.
$(a - 6)^2 + 4^2 = 16 \implies (a - 6)^2 + 16 = 16 \implies (a - 6)^2 = 0 \implies a = 6$.
Next,since $OC^2 = OB^2$,we have $(10 - 6)^2 + (b - (-5))^2 = 16$.
$4^2 + (b + 5)^2 = 16 \implies 16 + (b + 5)^2 = 16 \implies (b + 5)^2 = 0 \implies b = -5$.
Thus,the values are $a = 6$ and $b = -5$.

(D) The circumcentre $P(2, 3)$ is equidistant from all vertices of the triangle. Therefore, $PA = PB = PC$.
First, calculate the distance $PB$:
$PB^2 = (5 - 2)^2 + (1 - 3)^2 = 3^2 + (-2)^2 = 9 + 4 = 13$.
Since $PA^2 = PB^2$, we have:
$(a - 2)^2 + (6 - 3)^2 = 13$
$(a - 2)^2 + 3^2 = 13$
$(a - 2)^2 + 9 = 13$
$(a - 2)^2 = 4$
$a - 2 = \pm 2$
This gives two possible values for $a$: $a = 2 + 2 = 4$ or $a = 2 - 2 = 0$.
If $a = 4$, the coordinates of $A$ are $(4, 6)$, which is the same as the coordinates of $C(4, 6)$. Since $A$ and $C$ would coincide, a triangle cannot be formed.
Therefore, the only valid value is $a = 0$.

(C) Let the point on the $X-$axis be $P(x, 0)$.
The distance between $P(x, 0)$ and $A(4, 4)$ is given as $5$ units.
Using the distance formula: $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = d$
$\sqrt{(x - 4)^2 + (0 - 4)^2} = 5$
Squaring both sides: $(x - 4)^2 + (-4)^2 = 5^2$
$(x - 4)^2 + 16 = 25$
$(x - 4)^2 = 9$
Taking the square root: $x - 4 = \pm 3$
Case $1$: $x - 4 = 3 \implies x = 7$
Case $2$: $x - 4 = -3 \implies x = 1$
Therefore,the coordinates of the points are $(1, 0)$ and $(7, 0)$.

(N/A) To show that the points $A(1, 4)$,$B(7, -2)$,and $C(9, 6)$ form an isosceles triangle,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of $AB$:
$AB = \sqrt{(7 - 1)^2 + (-2 - 4)^2} = \sqrt{6^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$ units.
$2$. Length of $BC$:
$BC = \sqrt{(9 - 7)^2 + (6 - (-2))^2} = \sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17}$ units.
$3$. Length of $AC$:
$AC = \sqrt{(9 - 1)^2 + (6 - 4)^2} = \sqrt{8^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17}$ units.
Since $BC = AC$,two sides of the triangle are equal in length. Therefore,$\triangle ABC$ is an isosceles triangle.

(N/A) Let the points be $A(4, 3)$,$B(5, 1)$,and $C(1, 9)$.
Three points are collinear if the sum of the lengths of any two segments is equal to the length of the third segment.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$1$. Distance $AB = \sqrt{(5 - 4)^2 + (1 - 3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.
$2$. Distance $BC = \sqrt{(1 - 5)^2 + (9 - 1)^2} = \sqrt{(-4)^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$.
$3$. Distance $AC = \sqrt{(1 - 4)^2 + (9 - 3)^2} = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$.
Since $AB + AC = \sqrt{5} + 3\sqrt{5} = 4\sqrt{5} = BC$,the points $A$,$B$,and $C$ are collinear.

(N/A) Let the given points be $A(-2, 1)$,$B(2, -2)$,and $C(5, 2)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB^2 = (2 - (-2))^2 + (-2 - 1)^2 = (4)^2 + (-3)^2 = 16 + 9 = 25$.
$BC^2 = (5 - 2)^2 + (2 - (-2))^2 = (3)^2 + (4)^2 = 9 + 16 = 25$.
$AC^2 = (5 - (-2))^2 + (2 - 1)^2 = (7)^2 + (1)^2 = 49 + 1 = 50$.
We observe that $AB^2 + BC^2 = 25 + 25 = 50 = AC^2$.
Since the sum of the squares of two sides is equal to the square of the third side,by the converse of the Pythagoras theorem,$\triangle ABC$ is a right-angled triangle with the right angle at vertex $B$.

(C) Let the point on the $X$-axis be $P(x, 0)$.
Since $P$ is equidistant from $A(5, 4)$ and $B(-2, 3)$,we have $PA = PB$.
Using the distance formula,$PA^2 = PB^2$.
$(x - 5)^2 + (0 - 4)^2 = (x - (-2))^2 + (0 - 3)^2$
$(x - 5)^2 + 16 = (x + 2)^2 + 9$
$x^2 - 10x + 25 + 16 = x^2 + 4x + 4 + 9$
$-10x + 41 = 4x + 13$
$41 - 13 = 4x + 10x$
$28 = 14x$
$x = 2$.
Therefore,the point is $(2, 0)$.

(A) Let the point on the $Y$-axis be $P(0, y)$.
Since $P$ is equidistant from $A(3, 1)$ and $B(-2, 5)$, we have $PA = PB$, which implies $PA^2 = PB^2$.
Using the distance formula: $(3 - 0)^2 + (1 - y)^2 = (-2 - 0)^2 + (5 - y)^2$.
$9 + (1 - 2y + y^2) = 4 + (25 - 10y + y^2)$.
$10 - 2y + y^2 = 29 - 10y + y^2$.
Subtracting $y^2$ from both sides: $10 - 2y = 29 - 10y$.
Rearranging terms: $10y - 2y = 29 - 10$.
$8y = 19$.
$y = \frac{19}{8}$.
The point is $(0, \frac{19}{8})$.

(N/A) Let the vertices of the triangle be $A(10, -18), B(3, 6)$ and $C(-5, 2)$.
To determine if the triangle is isosceles,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of side $AB$:
$AB = \sqrt{(3 - 10)^2 + (6 - (-18))^2} = \sqrt{(-7)^2 + (24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ units.
$2$. Length of side $BC$:
$BC = \sqrt{(-5 - 3)^2 + (2 - 6)^2} = \sqrt{(-8)^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$ units.
$3$. Length of side $AC$:
$AC = \sqrt{(-5 - 10)^2 + (2 - (-18))^2} = \sqrt{(-15)^2 + (20)^2} = \sqrt{225 + 400} = \sqrt{625} = 25$ units.
Since the lengths of sides $AB$ and $AC$ are equal ($AB = AC = 25$ units),the triangle is an isosceles triangle.

(D) Let the vertices of the triangle be $A(4,6), B(5,1),$ and $C(4,0)$. Let the circumcentre be $P(x,y)$.
Since the circumcentre is equidistant from all vertices,$PA^2 = PB^2 = PC^2$.
First,equate $PA^2 = PC^2$:
$(x-4)^2 + (y-6)^2 = (x-4)^2 + (y-0)^2$
$(y-6)^2 = y^2$
$y^2 - 12y + 36 = y^2$
$12y = 36 \implies y = 3$.
Now,equate $PA^2 = PB^2$ with $y=3$:
$(x-4)^2 + (3-6)^2 = (x-5)^2 + (3-1)^2$
$(x-4)^2 + (-3)^2 = (x-5)^2 + (2)^2$
$x^2 - 8x + 16 + 9 = x^2 - 10x + 25 + 4$
$-8x + 25 = -10x + 29$
$2x = 4 \implies x = 2$.
Thus,the circumcentre is $(2,3)$.

(N/A) Let the vertices be $A(7,3)$,$B(3,0)$,$C(0,-4)$,and $D(4,-1)$.
To show that $ABCD$ is a rhombus,we must prove that all four sides are equal in length $(AB = BC = CD = DA)$ and the diagonals are not equal $(AC \neq BD)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(3 - 7)^2 + (0 - 3)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
$BC = \sqrt{(0 - 3)^2 + (-4 - 0)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
$CD = \sqrt{(4 - 0)^2 + (-1 - (-4))^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
$DA = \sqrt{(7 - 4)^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Since $AB = BC = CD = DA = 5$,all sides are equal.
Now,calculate the diagonals:
$AC = \sqrt{(0 - 7)^2 + (-4 - 3)^2} = \sqrt{(-7)^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2}$.
$BD = \sqrt{(4 - 3)^2 + (-1 - 0)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
Since $AC \neq BD$,the diagonals are not equal.
Therefore,$ABCD$ is a rhombus.

(N/A) Given coordinates are $A(b, c)$,$B(-a, 0)$,and $C(a, 0)$.
Since $D$ is the midpoint of $\overline{BC}$,the coordinates of $D$ are $(\frac{-a+a}{2}, \frac{0+0}{2}) = (0, 0)$.
Now,calculate the squares of the lengths:
$AB^2 = (b - (-a))^2 + (c - 0)^2 = (b+a)^2 + c^2 = b^2 + 2ab + a^2 + c^2$.
$AC^2 = (b - a)^2 + (c - 0)^2 = b^2 - 2ab + a^2 + c^2$.
Adding them: $AB^2 + AC^2 = (b^2 + 2ab + a^2 + c^2) + (b^2 - 2ab + a^2 + c^2) = 2(a^2 + b^2 + c^2)$.
Now calculate the right side:
$AD^2 = (b - 0)^2 + (c - 0)^2 = b^2 + c^2$.
$BD^2 = (-a - 0)^2 + (0 - 0)^2 = a^2$.
Thus,$2(AD^2 + BD^2) = 2(b^2 + c^2 + a^2)$.
Since both sides are equal to $2(a^2 + b^2 + c^2)$,the identity is proven.

(C) Let the third vertex be $C(x, y)$. Since $ABC$ is an equilateral triangle,$AB = BC = AC$.
The distance $AB = \sqrt{(1-0)^2 + (\sqrt{3}-0)^2} = \sqrt{1 + 3} = 2$.
Since $AB = BC = AC = 2$,we have:
$x^2 + y^2 = 2^2 = 4$ ---$(1)$
$(x-1)^2 + (y-\sqrt{3})^2 = 4$ ---$(2)$
Expanding $(2)$: $x^2 - 2x + 1 + y^2 - 2\sqrt{3}y + 3 = 4$.
Substituting $x^2 + y^2 = 4$ into the equation: $4 - 2x + 1 - 2\sqrt{3}y + 3 = 4$,which simplifies to $2x + 2\sqrt{3}y = 4$ or $x + \sqrt{3}y = 2$.
Thus,$x = 2 - \sqrt{3}y$.
Substituting this into $(1)$: $(2 - \sqrt{3}y)^2 + y^2 = 4$.
$4 - 4\sqrt{3}y + 3y^2 + y^2 = 4$,so $4y^2 - 4\sqrt{3}y = 0$.
$4y(y - \sqrt{3}) = 0$,giving $y = 0$ or $y = \sqrt{3}$.
If $y = 0$,$x = 2 - 0 = 2$. If $y = \sqrt{3}$,$x = 2 - \sqrt{3}(\sqrt{3}) = 2 - 3 = -1$.
Therefore,the third vertex $C$ can be $(2, 0)$ or $(-1, \sqrt{3})$.

(D) To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$,we use the distance formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Given points are $(6, 8)$ and $(3, 4)$.
Here,$x_1 = 6, y_1 = 8$ and $x_2 = 3, y_2 = 4$.
Substituting these values into the formula:
$d = \sqrt{(3 - 6)^2 + (4 - 8)^2}$
$d = \sqrt{(-3)^2 + (-4)^2}$
$d = \sqrt{9 + 16}$
$d = \sqrt{25}$
$d = 5$
Therefore,the distance between the points is $5$.

(A) To find the distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$,we use the distance formula:
$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Given $P(4, -7)$ and $Q(-1, 5)$,we have $x_1 = 4, y_1 = -7, x_2 = -1, y_2 = 5$.
Substituting these values into the formula:
$PQ = \sqrt{(-1 - 4)^2 + (5 - (-7))^2}$
$PQ = \sqrt{(-5)^2 + (5 + 7)^2}$
$PQ = \sqrt{(-5)^2 + (12)^2}$
$PQ = \sqrt{25 + 144}$
$PQ = \sqrt{169}$
$PQ = 13$
Therefore,the distance $PQ$ is $13$.

(B) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $(2, -3)$ and $(5, b)$ with distance $d = 5$.
Substituting the values into the formula:
$5 = \sqrt{(5 - 2)^2 + (b - (-3))^2}$
$5 = \sqrt{(3)^2 + (b + 3)^2}$
Squaring both sides:
$25 = 9 + (b + 3)^2$
$16 = (b + 3)^2$
Taking the square root on both sides:
$b + 3 = \pm 4$
Case $1$: $b + 3 = 4 \implies b = 1$
Case $2$: $b + 3 = -4 \implies b = -7$
Since $1$ is one of the given options,the correct value is $b = 1$.

(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $(a, 1)$ and $(8, a)$ with distance $d = 5$.
Substituting the values into the formula:
$5 = \sqrt{(8 - a)^2 + (a - 1)^2}$
Squaring both sides:
$25 = (8 - a)^2 + (a - 1)^2$
$25 = (64 - 16a + a^2) + (a^2 - 2a + 1)$
$25 = 2a^2 - 18a + 65$
$2a^2 - 18a + 40 = 0$
Dividing by $2$:
$a^2 - 9a + 20 = 0$
Factoring the quadratic equation:
$(a - 4)(a - 5) = 0$
Thus,$a = 4$ or $a = 5$.
Since $5$ is one of the given options,the correct answer is $5$.

(C) To determine the type of triangle $\Delta PQR$,we calculate the lengths of its sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of $PQ$: $PQ = \sqrt{(0 - 0)^2 + (5 - 0)^2} = \sqrt{0^2 + 5^2} = 5 \text{ units}$.
$2$. Length of $QR$: $QR = \sqrt{(6 - 0)^2 + (0 - 5)^2} = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61} \text{ units}$.
$3$. Length of $PR$: $PR = \sqrt{(6 - 0)^2 + (0 - 0)^2} = \sqrt{6^2 + 0^2} = 6 \text{ units}$.
Since the vertices $P(0,0), Q(0,5),$ and $R(6,0)$ lie on the axes,the angle at the origin $P(0,0)$ is $90^\circ$ because the $y$-axis $(PQ)$ is perpendicular to the $x$-axis $(PR)$.
Since one angle is $90^\circ$,$\Delta PQR$ is a right-angled triangle.

(A) To check if the points $A(-1, 4), B(2, 3),$ and $C(8, 1)$ are collinear,we calculate the slopes of the segments $AB$ and $BC$.
Slope of $AB = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 4}{2 - (-1)} = \frac{-1}{3}$.
Slope of $BC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 3}{8 - 2} = \frac{-2}{6} = \frac{-1}{3}$.
Since the slope of $AB$ is equal to the slope of $BC$,the points $A, B,$ and $C$ are collinear.
Now,we check the distances between the points:
$AB = \sqrt{(2 - (-1))^2 + (3 - 4)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.
$BC = \sqrt{(8 - 2)^2 + (1 - 3)^2} = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$.
$AC = \sqrt{(8 - (-1))^2 + (1 - 4)^2} = \sqrt{9^2 + (-3)^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10}$.
Since $AB + BC = \sqrt{10} + 2\sqrt{10} = 3\sqrt{10} = AC$,the point $B$ lies between $A$ and $C$.
Therefore,the order of the points is $A-B-C$.

(B) The diameter of the circle is the distance between the points $(2, 4)$ and $(2, 6)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, we get:
Diameter $= \sqrt{(2 - 2)^2 + (6 - 4)^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2$.
The radius $r$ is half of the diameter.
$r = \frac{\text{Diameter}}{2} = \frac{2}{2} = 1$.
Therefore, the radius of the circle is $1$.

(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given points are $(5, 0)$ and $(x, 4)$ with distance $d = \sqrt{17}$.
Substituting the values: $\sqrt{17} = \sqrt{(x - 5)^2 + (4 - 0)^2}$.
Squaring both sides: $17 = (x - 5)^2 + 4^2$.
$17 = (x - 5)^2 + 16$.
$(x - 5)^2 = 17 - 16 = 1$.
Taking the square root: $x - 5 = \pm 1$.
Case $1$: $x - 5 = 1 \implies x = 6$.
Case $2$: $x - 5 = -1 \implies x = 4$.
Thus,$x = 4, 6$.

(C) The coordinates of a point are given as $(x, y)$.
Here,the point is $(6, 2)$,where $x = 6$ and $y = 2$.
The perpendicular distance of any point $(x, y)$ from the $X-$ axis is given by the absolute value of its $y-$ coordinate,which is $|y|$.
Therefore,the perpendicular distance of the point $(6, 2)$ from the $X-$ axis is $|2| = 2$ units.

(D) The coordinates of the given point are $(x, y) = (-3, 8)$.
In a Cartesian coordinate system,the perpendicular distance of any point $(x, y)$ from the $Y$-axis is given by the absolute value of its $x$-coordinate,i.e.,$|x|$.
Here,the $x$-coordinate is $-3$.
Therefore,the perpendicular distance from the $Y$-axis is $|-3| = 3$.
Thus,the correct option is $D$.

(A) The coordinates of point $B$ are $(0, 12)$ and the coordinates of point $C$ are $(5, 0)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$,we calculate the distance $BC$:
$BC = \sqrt{(5 - 0)^2 + (0 - 12)^2}$
$BC = \sqrt{(5)^2 + (-12)^2}$
$BC = \sqrt{25 + 144}$
$BC = \sqrt{169}$
$BC = 13$
Thus,the distance $BC$ is $13$ units.

(D) Let the vertices be $A(1, 1)$, $B(-1, -1)$, and $C(-\sqrt{3}, \sqrt{3})$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(-1 - 1)^2 + (-1 - 1)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.
$BC = \sqrt{(-\sqrt{3} - (-1))^2 + (\sqrt{3} - (-1))^2} = \sqrt{(1 - \sqrt{3})^2 + (\sqrt{3} + 1)^2} = \sqrt{(1 - 2\sqrt{3} + 3) + (3 + 2\sqrt{3} + 1)} = \sqrt{4 - 2\sqrt{3} + 4 + 2\sqrt{3}} = \sqrt{8} = 2\sqrt{2}$.
$CA = \sqrt{(1 - (-\sqrt{3}))^2 + (1 - \sqrt{3})^2} = \sqrt{(1 + \sqrt{3})^2 + (1 - \sqrt{3})^2} = \sqrt{(1 + 2\sqrt{3} + 3) + (1 - 2\sqrt{3} + 3)} = \sqrt{4 + 2\sqrt{3} + 4 - 2\sqrt{3}} = \sqrt{8} = 2\sqrt{2}$.
Since $AB = BC = CA = 2\sqrt{2}$, all three sides are equal.
Therefore, the triangle is an equilateral triangle.

(D) Let the vertices of the triangle be $A(0,0)$,$B(4.2,0)$,and $C(0,9.1)$.
Since vertex $A$ lies at the origin $(0,0)$,side $AB$ lies along the $x$-axis and side $AC$ lies along the $y$-axis.
The angle between the $x$-axis and the $y$-axis is $90^\circ$.
Therefore,the angle $\angle BAC = 90^\circ$.
Since one angle of the triangle is $90^\circ$,it is a right-angled triangle.

(B) The distance formula between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by $d = \sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}$.
Given points are $P(x_{1}, 0)$ and $Q(x_{2}, 0)$.
Substituting these values into the distance formula:
$PQ = \sqrt{(x_{2}-x_{1})^2 + (0-0)^2}$
$PQ = \sqrt{(x_{2}-x_{1})^2}$
Since the distance must always be non-negative,we take the absolute value:
$PQ = |x_{2}-x_{1}| = |x_{1}-x_{2}|$.
Therefore,the correct option is $B$.

(C) The distance formula between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$.
Given the points $A(0, y_{1})$ and $B(0, y_{2})$,we substitute these coordinates into the formula:
$AB = \sqrt{(0 - 0)^{2} + (y_{2} - y_{1})^{2}}$
$AB = \sqrt{0^{2} + (y_{2} - y_{1})^{2}}$
$AB = \sqrt{(y_{2} - y_{1})^{2}}$
Since the distance must be non-negative,we take the absolute value:
$AB = |y_{2} - y_{1}|$ or $|y_{1} - y_{2}|$.
Therefore,the correct option is $C$.