Problems related to triangle and quadrilateral Questions in English

(C) Let the sides be $a = 3x + 4y$,$b = 4x + 3y$,and $c = 5x + 5y$.
Since $x, y > 0$,we compare the sides. Note that $c = 5x + 5y$ is the largest side because $5x + 5y > 3x + 4y$ and $5x + 5y > 4x + 3y$.
For a triangle to be obtuse,the square of the largest side must be greater than the sum of the squares of the other two sides $(c^2 > a^2 + b^2)$.
$a^2 + b^2 = (3x + 4y)^2 + (4x + 3y)^2 = (9x^2 + 24xy + 16y^2) + (16x^2 + 24xy + 9y^2) = 25x^2 + 48xy + 25y^2$.
$c^2 = (5x + 5y)^2 = 25x^2 + 50xy + 25y^2$.
Comparing $c^2$ and $a^2 + b^2$,we see that $25x^2 + 50xy + 25y^2 > 25x^2 + 48xy + 25y^2$ since $x, y > 0$.
Thus,$c^2 > a^2 + b^2$,which implies the triangle is obtuse angled.

(C) The vertices of the square are determined by the origin $(0, 0)$ and the side length of $5$ units along the positive axes.
Since the sides lie along the positive $x$ and $y$ axes,the vertices are:
$1$. Origin: $(0, 0)$
$2$. Point on $x$-axis: $(5, 0)$
$3$. Point on $y$-axis: $(0, 5)$
$4$. Opposite vertex: $(5, 5)$
Comparing these with the given options,$(-5, -5)$ is not a vertex of the square because the square lies entirely in the first quadrant.

(C) Let the vertices be $O(0, 0)$,$A(2, 2\sqrt{3})$,and $B(a, b)$.
Since it is an equilateral triangle,the side length $l$ is the distance $OA = \sqrt{(2-0)^2 + (2\sqrt{3}-0)^2} = \sqrt{4 + 12} = \sqrt{16} = 4$.
Thus,$OB^2 = a^2 + b^2 = 4^2 = 16$ (Equation $1$).
Also,$AB^2 = (a-2)^2 + (b-2\sqrt{3})^2 = 16$.
Expanding this: $a^2 - 4a + 4 + b^2 - 4\sqrt{3}b + 12 = 16$.
Substituting $a^2 + b^2 = 16$: $16 - 4a - 4\sqrt{3}b + 16 = 16$,which simplifies to $4a + 4\sqrt{3}b = 16$,or $a + \sqrt{3}b = 4$ (Equation $2$).
From Equation $2$,$a = 4 - \sqrt{3}b$. Substituting into Equation $1$: $(4 - \sqrt{3}b)^2 + b^2 = 16$.
$16 - 8\sqrt{3}b + 3b^2 + b^2 = 16 \Rightarrow 4b^2 - 8\sqrt{3}b = 0$.
$4b(b - 2\sqrt{3}) = 0$,so $b = 0$ or $b = 2\sqrt{3}$.
If $b = 0$,$a = 4$. If $b = 2\sqrt{3}$,$a = 4 - 6 = -2$.
Since $(a, b)$ must be distinct from $(2, 2\sqrt{3})$,the solution is $(4, 0)$.

(C) Let the vertices of the equilateral triangle be $O(0, 0)$,$A(4, 0)$,and $B(x, y)$.
Since the triangle is equilateral,the side length $s = \sqrt{(4-0)^2 + (0-0)^2} = 4$.
The distance from $O(0, 0)$ to $B(x, y)$ must be $4$,so $x^2 + y^2 = 4^2 = 16$.
The distance from $A(4, 0)$ to $B(x, y)$ must also be $4$,so $(x-4)^2 + y^2 = 4^2 = 16$.
Expanding the second equation: $x^2 - 8x + 16 + y^2 = 16$.
Substitute $x^2 + y^2 = 16$ into the equation: $16 - 8x + 16 = 16$,which simplifies to $8x = 16$,so $x = 2$.
Substitute $x = 2$ into $x^2 + y^2 = 16$: $2^2 + y^2 = 16 \implies 4 + y^2 = 16 \implies y^2 = 12 \implies y = \pm \sqrt{12} = \pm 2\sqrt{3}$.
Thus,the third vertex is $(2, \pm 2\sqrt{3})$.

(D) In $\Delta ABC$,the vertices are $A \equiv (-3, 0)$,$B \equiv (4, -1)$,and $C \equiv (5, 2)$.
First,we calculate the length of the base $BC$ using the distance formula:
$BC = \sqrt{(5 - 4)^2 + (2 - (-1))^2} = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$.
Next,we calculate the area of $\Delta ABC$ using the coordinate formula:
$\text{Area} = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$
$= \frac{1}{2} |(-3)(-1 - 2) + 4(2 - 0) + 5(0 - (-1))|$
$= \frac{1}{2} |(-3)(-3) + 4(2) + 5(1)|$
$= \frac{1}{2} |9 + 8 + 5| = \frac{1}{2} |22| = 11$.
Since the area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{altitude}$,we have:
$11 = \frac{1}{2} \times \sqrt{10} \times \text{altitude}$
$\text{altitude} = \frac{22}{\sqrt{10}}$.

(C) Let the vertices be $A(1, 1)$,$B(-1, -1)$,and $C(-\sqrt{3}, k)$.
For an equilateral triangle,the distance between any two points must be equal.
First,calculate the length of side $AB$:
$AB^2 = (-1 - 1)^2 + (-1 - 1)^2 = (-2)^2 + (-2)^2 = 4 + 4 = 8$.
Now,calculate the distance $AC^2$:
$AC^2 = (-\sqrt{3} - 1)^2 + (k - 1)^2 = (\sqrt{3} + 1)^2 + (k - 1)^2 = 3 + 1 + 2\sqrt{3} + (k - 1)^2 = 4 + 2\sqrt{3} + (k - 1)^2$.
Since $AB^2 = AC^2$,we have:
$8 = 4 + 2\sqrt{3} + (k - 1)^2$
$(k - 1)^2 = 4 - 2\sqrt{3} = (\sqrt{3} - 1)^2$.
Taking the square root,$k - 1 = \pm(\sqrt{3} - 1)$.
If $k - 1 = \sqrt{3} - 1$,then $k = \sqrt{3}$.
If $k - 1 = -(\sqrt{3} - 1) = 1 - \sqrt{3}$,then $k = 2 - \sqrt{3}$.
Comparing with the given options,the correct value is $k = \sqrt{3}$.

(A) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.
Given mid-points are $(2, 1)$,$(-1, -3)$,and $(4, 5)$.
For $x$-coordinates: $\frac{x_1 + x_2}{2} = 2$,$\frac{x_2 + x_3}{2} = -1$,$\frac{x_3 + x_1}{2} = 4$.
Adding these equations: $x_1 + x_2 + x_3 = 5$.
Subtracting individual equations: $x_3 = 5 - 4 = 1$,$x_1 = 5 - (-2) = 7$,$x_2 = 5 - 8 = -3$.
For $y$-coordinates: $\frac{y_1 + y_2}{2} = 1$,$\frac{y_2 + y_3}{2} = -3$,$\frac{y_3 + y_1}{2} = 5$.
Adding these equations: $y_1 + y_2 + y_3 = 3$.
Subtracting individual equations: $y_3 = 3 - 2 = 1$,$y_1 = 3 - (-6) = 9$,$y_2 = 3 - 10 = -7$.
The vertices are $(7, 9), (-3, -7)$,and $(1, 1)$.

(C) Let the vertices of the rectangle be $A(2, -2)$,$B(8, 4)$,$C(5, 7)$,and $D(x, y)$.
Since the diagonals of a rectangle bisect each other,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.
Midpoint of $AC = (\frac{2 + 5}{2}, \frac{-2 + 7}{2}) = (3.5, 2.5)$.
Midpoint of $BD = (\frac{8 + x}{2}, \frac{4 + y}{2})$.
Equating the midpoints:
$\frac{8 + x}{2} = 3.5$ $\Rightarrow 8 + x = 7$ $\Rightarrow x = -1$.
$\frac{4 + y}{2} = 2.5$ $\Rightarrow 4 + y = 5$ $\Rightarrow y = 1$.
Thus,the fourth vertex is $(-1, 1)$.

(C) In a parallelogram,the diagonals bisect each other at their midpoint.
Let the midpoint of diagonal $PR$ be $M_1$ and the midpoint of diagonal $QS$ be $M_2$.
Since $M_1 = M_2$,we have:
Midpoint of $PR = (\frac{1+5}{2}, \frac{2+7}{2}) = (3, 4.5)$
Midpoint of $QS = (\frac{4+a}{2}, \frac{6+b}{2})$
Equating the coordinates:
$\frac{4+a}{2} = 3$ $\Rightarrow 4+a = 6$ $\Rightarrow a = 2$
$\frac{6+b}{2} = 4.5$ $\Rightarrow 6+b = 9$ $\Rightarrow b = 3$
Thus,$a = 2$ and $b = 3$.

(C) Let the given vertices be $A(0, -1)$ and $C(0, 3)$.
The length of the diagonal $AC = \sqrt{(0-0)^2 + (3 - (-1))^2} = \sqrt{0^2 + 4^2} = 4$.
Let the other two vertices be $B(x, y)$ and $D(x', y')$.
In a square,the diagonals bisect each other at right angles and are equal in length.
The midpoint of $AC$ is $(\frac{0+0}{2}, \frac{-1+3}{2}) = (0, 1)$.
Since the diagonals bisect each other,the midpoint of $BD$ is also $(0, 1)$.
Thus,$\frac{x+x'}{2} = 0 \Rightarrow x' = -x$ and $\frac{y+y'}{2} = 1 \Rightarrow y' = 2-y$.
Also,the distance from the midpoint $(0, 1)$ to any vertex is half the diagonal length,which is $2$.
For vertex $B(x, y)$,the distance to $(0, 1)$ is $\sqrt{(x-0)^2 + (y-1)^2} = 2$,so $x^2 + (y-1)^2 = 4$.
Since $AB \perp BC$,the slope of $AB$ is $\frac{y+1}{x-0} = \frac{y+1}{x}$ and the slope of $BC$ is $\frac{y-3}{x-0} = \frac{y-3}{x}$.
Since $AB \perp BC$,their product is $-1$: $(\frac{y+1}{x})(\frac{y-3}{x}) = -1$ $\Rightarrow (y+1)(y-3) = -x^2$ $\Rightarrow y^2 - 2y - 3 = -x^2$ $\Rightarrow x^2 + y^2 - 2y - 3 = 0$.
Substituting $x^2 = 4 - (y-1)^2 = 4 - (y^2 - 2y + 1) = 3 + 2y - y^2$ into the equation:
$(3 + 2y - y^2) + y^2 - 2y - 3 = 0$,which is $0=0$.
This confirms the midpoint logic. Since $AC$ is vertical,$BD$ must be horizontal,so $y = 1$.
Substituting $y=1$ into $x^2 + (y-1)^2 = 4$,we get $x^2 + 0 = 4$,so $x = \pm 2$.
Thus,the other two vertices are $(2, 1)$ and $(-2, 1)$.

(C) Let the vertices be $A(-2, 2)$,$B(8, -2)$,and $C(-4, -3)$.
Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB^2 = (8 - (-2))^2 + (-2 - 2)^2 = 10^2 + (-4)^2 = 100 + 16 = 116$
$BC^2 = (-4 - 8)^2 + (-3 - (-2))^2 = (-12)^2 + (-1)^2 = 144 + 1 = 145$
$AC^2 = (-4 - (-2))^2 + (-3 - 2)^2 = (-2)^2 + (-5)^2 = 4 + 25 = 29$
Observe that $AB^2 + AC^2 = 116 + 29 = 145 = BC^2$.
Since the sum of the squares of two sides equals the square of the third side,it is a right-angled triangle.

(B) Let $A = \left( \frac{a}{\sqrt{3}}, a \right)$,$B = \left( \frac{2a}{\sqrt{3}}, 2a \right)$,and $C = \left( \frac{a}{\sqrt{3}}, 3a \right)$.
Calculate the squared distances between the vertices:
$AB^2 = \left( \frac{a}{\sqrt{3}} - \frac{2a}{\sqrt{3}} \right)^2 + (a - 2a)^2 = \left( -\frac{a}{\sqrt{3}} \right)^2 + (-a)^2 = \frac{a^2}{3} + a^2 = \frac{4a^2}{3}$.
$BC^2 = \left( \frac{2a}{\sqrt{3}} - \frac{a}{\sqrt{3}} \right)^2 + (2a - 3a)^2 = \left( \frac{a}{\sqrt{3}} \right)^2 + (-a)^2 = \frac{a^2}{3} + a^2 = \frac{4a^2}{3}$.
$AC^2 = \left( \frac{a}{\sqrt{3}} - \frac{a}{\sqrt{3}} \right)^2 + (a - 3a)^2 = 0^2 + (-2a)^2 = 4a^2$.
Since $AB^2 = BC^2 = \frac{4a^2}{3}$,the sides $AB$ and $BC$ are equal in length.
Therefore,the triangle is an isosceles triangle.

(D) Let the points be $A(0, 8/3)$,$B(1, 3)$,and $C(82, 30)$.
To check if these points form a triangle,we calculate the area of the triangle using the formula:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the values:
$\text{Area} = \frac{1}{2} |0(3 - 30) + 1(30 - 8/3) + 82(8/3 - 3)|$
$\text{Area} = \frac{1}{2} |0 + (90 - 8)/3 + 82(-1/3)|$
$\text{Area} = \frac{1}{2} |82/3 - 82/3| = 0$
Since the area is $0$,the points are collinear and do not form a triangle. Therefore,the correct option is $D$.

(B) Let the vertices be $A(-1, 1)$,$B(0, -3)$,$C(5, 2)$,and $D(4, 6)$.
Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(0 - (-1))^2 + (-3 - 1)^2} = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}$
$BC = \sqrt{(5 - 0)^2 + (2 - (-3))^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50}$
$CD = \sqrt{(4 - 5)^2 + (6 - 2)^2} = \sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}$
$DA = \sqrt{(-1 - 4)^2 + (1 - 6)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50}$
Since opposite sides are equal ($AB = CD = \sqrt{17}$ and $BC = DA = \sqrt{50}$),the quadrilateral is a parallelogram.
Check the diagonals:
$AC = \sqrt{(5 - (-1))^2 + (2 - 1)^2} = \sqrt{6^2 + 1^2} = \sqrt{37}$
$BD = \sqrt{(4 - 0)^2 + (6 - (-3))^2} = \sqrt{4^2 + 9^2} = \sqrt{16 + 81} = \sqrt{97}$
Since $AC \neq BD$,it is not a rectangle or a square.
Thus,the quadrilateral is a parallelogram.

(B) Let the vertices be $A(-4, -1)$,$B(-2, -4)$,$C(4, 0)$,and $D(2, 3)$.
First,we check the mid-points of the diagonals $AC$ and $BD$:
Mid-point of $AC = (\frac{-4+4}{2}, \frac{-1+0}{2}) = (0, -0.5)$.
Mid-point of $BD = (\frac{-2+2}{2}, \frac{-4+3}{2}) = (0, -0.5)$.
Since the mid-points are the same,the figure is a parallelogram.
Next,we calculate the slopes of adjacent sides $AB$ and $AD$:
Slope of $AB$ $(m_1)$ = $\frac{-4 - (-1)}{-2 - (-4)} = \frac{-3}{2}$.
Slope of $AD$ $(m_2)$ = $\frac{3 - (-1)}{2 - (-4)} = \frac{4}{6} = \frac{2}{3}$.
Since $m_1 \times m_2 = (\frac{-3}{2}) \times (\frac{2}{3}) = -1$,the adjacent sides are perpendicular.
Therefore,the figure is a rectangle.

(D) Let the vertices be $A(0, 2)$,$B(1, 0)$,and $C(3, 1)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(1 - 0)^2 + (0 - 2)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$
$BC = \sqrt{(3 - 1)^2 + (1 - 0)^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$
$AC = \sqrt{(3 - 0)^2 + (1 - 2)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$
Since $AB = BC = \sqrt{5}$,the triangle is isosceles.
Also,$(AB)^2 + (BC)^2 = 5 + 5 = 10$ and $(AC)^2 = 10$.
Since $(AB)^2 + (BC)^2 = (AC)^2$,the triangle is a right-angled triangle.
Therefore,the triangle is an isosceles right-angled triangle.

(B) Let $A = (0, 0)$,$B = (a, 0)$,and $C = \left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(a - 0)^2 + (0 - 0)^2} = \sqrt{a^2} = a$.
$BC = \sqrt{\left( \frac{a}{2} - a \right)^2 + \left( \frac{a\sqrt{3}}{2} - 0 \right)^2} = \sqrt{\left( -\frac{a}{2} \right)^2 + \left( \frac{a\sqrt{3}}{2} \right)^2} = \sqrt{\frac{a^2}{4} + \frac{3a^2}{4}} = \sqrt{a^2} = a$.
$AC = \sqrt{\left( \frac{a}{2} - 0 \right)^2 + \left( \frac{a\sqrt{3}}{2} - 0 \right)^2} = \sqrt{\frac{a^2}{4} + \frac{3a^2}{4}} = \sqrt{a^2} = a$.
Since $AB = BC = AC = a$,the triangle is an equilateral triangle.

(D) Calculate the side lengths using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
$AB = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9+16} = 5$.
$BC = \sqrt{(7-3)^2 + (7-4)^2} = \sqrt{4^2+3^2} = 5$.
$CD = \sqrt{(4-7)^2 + (3-7)^2} = \sqrt{(-3)^2+(-4)^2} = 5$.
$AD = \sqrt{(4-0)^2 + (3-0)^2} = \sqrt{4^2+3^2} = 5$.
Since all sides are equal,it is a rhombus or a square.
Calculate the diagonals:
$AC = \sqrt{(7-0)^2 + (7-0)^2} = \sqrt{49+49} = \sqrt{98} = 7\sqrt{2}$.
$BD = \sqrt{(4-3)^2 + (3-4)^2} = \sqrt{1^2+(-1)^2} = \sqrt{2}$.
Since $AC \neq BD$,the diagonals are not equal.
Therefore,$ABCD$ is a rhombus.

(B) Let the vertices be $A(0, -1), B(2, 1), C(0, 3),$ and $D(-2, 1).$
Calculate the side lengths using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(2-0)^2 + (1-(-1))^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$
$BC = \sqrt{(0-2)^2 + (3-1)^2} = \sqrt{4+4} = 2\sqrt{2}$
$CD = \sqrt{(-2-0)^2 + (1-3)^2} = \sqrt{4+4} = 2\sqrt{2}$
$DA = \sqrt{(0-(-2))^2 + (-1-1)^2} = \sqrt{4+4} = 2\sqrt{2}$
Calculate the diagonals:
$AC = \sqrt{(0-0)^2 + (3-(-1))^2} = \sqrt{0+16} = 4$
$BD = \sqrt{(-2-2)^2 + (1-1)^2} = \sqrt{16+0} = 4$
Since all sides are equal $(AB=BC=CD=DA)$ and diagonals are equal $(AC=BD)$,the quadrilateral is a square.

(A) Let the vertices be $A(4, 0)$,$B(-1, -1)$,and $C(3, 5)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(-1 - 4)^2 + (-1 - 0)^2} = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26}$.
$AC = \sqrt{(3 - 4)^2 + (5 - 0)^2} = \sqrt{(-1)^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}$.
$BC = \sqrt{(3 - (-1))^2 + (5 - (-1))^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}$.
Since $AB = AC = \sqrt{26}$,the triangle is isosceles.
Also,$AB^2 + AC^2 = 26 + 26 = 52$ and $BC^2 = 52$.
Since $AB^2 + AC^2 = BC^2$,the triangle is right-angled at $A$ by the converse of the Pythagorean theorem.
Thus,the triangle is isosceles and right-angled.

(B) To determine the type of triangle,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$PQ = \sqrt{(4 - 2)^2 + (-1 - 7)^2} = \sqrt{2^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68}$.
$PR = \sqrt{(-2 - 2)^2 + (6 - 7)^2} = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}$.
$QR = \sqrt{(-2 - 4)^2 + (6 - (-1))^2} = \sqrt{(-6)^2 + 7^2} = \sqrt{36 + 49} = \sqrt{85}$.
Now,check the squares of the sides:
$PQ^2 = 68$,$PR^2 = 17$,$QR^2 = 85$.
Since $PQ^2 + PR^2 = 68 + 17 = 85 = QR^2$,the triangle satisfies the Pythagorean theorem.
Therefore,it is a right-angled triangle.

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

(D) To determine which inequality is satisfied by the interior points of the triangle,we check the vertices $(1, 3)$,$(5, 0)$,and $(-1, 2)$ for each inequality.
For $3x + 2y \ge 0$:
$(1, 3) \implies 3(1) + 2(3) = 9 > 0$
$(5, 0) \implies 3(5) + 2(0) = 15 > 0$
$(-1, 2) \implies 3(-1) + 2(2) = 1 > 0$
Since all vertices satisfy the inequality,the interior points also satisfy it.
For $2x + y - 13 \le 0$:
$(1, 3) \implies 2(1) + 3 - 13 = -8 \le 0$
$(5, 0) \implies 2(5) + 0 - 13 = -3 \le 0$
$(-1, 2) \implies 2(-1) + 2 - 13 = -13 \le 0$
Since all vertices satisfy the inequality,the interior points also satisfy it.
For $2x - 3y - 12 \le 0$:
$(1, 3) \implies 2(1) - 3(3) - 12 = -19 \le 0$
$(5, 0) \implies 2(5) - 3(0) - 12 = -2 \le 0$
$(-1, 2) \implies 2(-1) - 3(2) - 12 = -20 \le 0$
Since all vertices satisfy the inequality,the interior points also satisfy it.
Therefore,all the given inequalities are satisfied.

(C) To find the vertices of the triangle,we solve the equations in pairs:
$1$. Intersection of $7x - 2y + 10 = 0$ and $y + 2 = 0$:
Substituting $y = -2$ into the first equation: $7x - 2(-2) + 10 = 0 \implies 7x + 4 + 10 = 0 \implies 7x = -14 \implies x = -2$. Vertex: $(-2, -2)$.
$2$. Intersection of $7x + 2y - 10 = 0$ and $y + 2 = 0$:
Substituting $y = -2$ into the second equation: $7x + 2(-2) - 10 = 0 \implies 7x - 4 - 10 = 0 \implies 7x = 14 \implies x = 2$. Vertex: $(2, -2)$.
$3$. Intersection of $7x - 2y + 10 = 0$ and $7x + 2y - 10 = 0$:
Adding the two equations: $14x = 0 \implies x = 0$. Substituting $x = 0$ into $7x - 2y + 10 = 0$: $-2y = -10 \implies y = 5$. Vertex: $(0, 5)$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area $= \frac{1}{2} |(-2)(-2 - 5) + 2(5 - (-2)) + 0(-2 - (-2))|$
Area $= \frac{1}{2} |(-2)(-7) + 2(7) + 0| = \frac{1}{2} |14 + 14| = \frac{28}{2} = 14 \, sq. \, units$.

(C) To find the vertices of the triangle,we solve the equations of the lines pairwise:
$1$. Intersection of $y = m_1x + c_1$ and $x = 0$ gives the point $(0, c_1)$.
$2$. Intersection of $y = m_2x + c_2$ and $x = 0$ gives the point $(0, c_2)$.
$3$. Intersection of $y = m_1x + c_1$ and $y = m_2x + c_2$ gives:
$m_1x + c_1 = m_2x + c_2 \implies x(m_1 - m_2) = c_2 - c_1 \implies x = \frac{c_2 - c_1}{m_1 - m_2}$.
Substituting $x$ into $y = m_1x + c_1$,we get $y = m_1(\frac{c_2 - c_1}{m_1 - m_2}) + c_1 = \frac{m_1c_2 - m_1c_1 + m_1c_1 - m_2c_1}{m_1 - m_2} = \frac{m_1c_2 - m_2c_1}{m_1 - m_2}$.
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the vertices $(0, c_1), (0, c_2), (\frac{c_2 - c_1}{m_1 - m_2}, \frac{m_1c_2 - m_2c_1}{m_1 - m_2})$:
Area $= \frac{1}{2} |0(c_2 - y_3) + 0(y_3 - c_1) + (\frac{c_2 - c_1}{m_1 - m_2})(c_1 - c_2)|$
Area $= \frac{1}{2} |\frac{-(c_1 - c_2)^2}{m_1 - m_2}| = \frac{1}{2} \frac{(c_1 - c_2)^2}{|m_1 - m_2|}$.

(A) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area of $\Delta ABC = \frac{1}{2} |6(5 - (-2)) + (-3)(-2 - 3) + 4(3 - 5)| = \frac{1}{2} |6(7) - 3(-5) + 4(-2)| = \frac{1}{2} |42 + 15 - 8| = \frac{1}{2} |49| = 24.5$.
Area of $\Delta PBC = \frac{1}{2} |x(5 - (-2)) + (-3)(-2 - y) + 4(y - 5)| = \frac{1}{2} |7x + 6 + 3y + 4y - 20| = \frac{1}{2} |7x + 7y - 14| = \frac{7}{2} |x + y - 2|$.
Ratio = $\frac{\text{Area}(\Delta PBC)}{\text{Area}(\Delta ABC)} = \frac{\frac{7}{2} |x + y - 2|}{\frac{49}{2}} = \left| \frac{x + y - 2}{7} \right|$.

(A) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
For $\Delta ABC$ with $A(6, 3)$,$B(-3, 5)$,$C(4, -2)$:
$\text{Area}(\Delta ABC) = \frac{1}{2} |6(5 - (-2)) + (-3)(-2 - 3) + 4(3 - 5)| = \frac{1}{2} |6(7) + (-3)(-5) + 4(-2)| = \frac{1}{2} |42 + 15 - 8| = \frac{49}{2}$.
For $\Delta DBC$ with $D(x, 3x)$,$B(-3, 5)$,$C(4, -2)$:
$\text{Area}(\Delta DBC) = \frac{1}{2} |x(5 - (-2)) + (-3)(-2 - 3x) + 4(3x - 5)| = \frac{1}{2} |7x + 6 + 9x + 12x - 20| = \frac{1}{2} |28x - 14| = |14x - 7|$.
Given $\frac{\text{Area}(\Delta DBC)}{\text{Area}(\Delta ABC)} = \frac{1}{2}$,we have $2 \times \text{Area}(\Delta DBC) = \text{Area}(\Delta ABC)$.
$2 |14x - 7| = \frac{49}{2} \Rightarrow |14x - 7| = \frac{49}{4}$.
Case $1$: $14x - 7 = \frac{49}{4}$ $\Rightarrow 14x = \frac{49}{4} + 7 = \frac{77}{4}$ $\Rightarrow x = \frac{77}{56} = \frac{11}{8}$.
Case $2$: $14x - 7 = -\frac{49}{4}$ $\Rightarrow 14x = 7 - \frac{49}{4} = -\frac{21}{4}$ $\Rightarrow x = -\frac{21}{56} = -\frac{3}{8}$.
Since $\frac{11}{8}$ is in the options,the correct value is $\frac{11}{8}$.

(B) The given lines are $x = 0$ (y-axis),$y = 0$ (x-axis),and $x + 2y + 3 = 0$.
To find the vertices of the triangle,we find the intersection points:
$1$. Intersection of $x = 0$ and $y = 0$ is $(0, 0)$.
$2$. Intersection of $x = 0$ and $x + 2y + 3 = 0$: Substituting $x = 0$,we get $2y = -3$,so $y = -1.5$. The point is $(0, -1.5)$.
$3$. Intersection of $y = 0$ and $x + 2y + 3 = 0$: Substituting $y = 0$,we get $x = -3$. The point is $(-3, 0)$.
The triangle is a right-angled triangle with base length $|-3| = 3$ and height $|-1.5| = 1.5$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 1.5 = \frac{4.5}{2} = 2.25$ sq. units.

(A) Three points $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are collinear if the area of the triangle formed by them is zero,or if the slope between any two pairs of points is equal.
Let the points be $P(a, b)$,$Q(a', b')$,and $R(a - a', b - b')$.
The slope of $PQ$ is $m_1 = \frac{b' - b}{a' - a}$.
The slope of $QR$ is $m_2 = \frac{(b - b') - b'}{(a - a') - a'} = \frac{b - 2b'}{a - 2a'}$.
Since the points are collinear,$m_1 = m_2$:
$\frac{b' - b}{a' - a} = \frac{b - 2b'}{a - 2a'}$
$(b' - b)(a - 2a') = (b - 2b')(a' - a)$
$ab' - 2a'b' - ab + 2a'b = a'b - ab - 2a'b' + 2ab'$
$ab' + 2a'b = a'b + 2ab'$
$a'b = ab'$
Therefore,$ab' = a'b$.

(C) For three points $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ to be collinear,the area of the triangle formed by them must be zero.
$\frac{1}{2} |a(b - 1) + 0(1 - 0) + 1(0 - b)| = 0$
$|ab - a - b| = 0$
$ab - a - b = 0$
$ab = a + b$
Dividing both sides by $ab$ (assuming $a, b \neq 0$):
$\frac{ab}{ab} = \frac{a}{ab} + \frac{b}{ab}$
$1 = \frac{1}{b} + \frac{1}{a}$
Therefore,$\frac{1}{a} + \frac{1}{b} = 1$.

(A) For three points $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ to be collinear,the area of the triangle formed by them must be zero.
The area of the triangle is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$.
Substituting the given points $(-5, 1), (p, 5)$ and $(10, 7)$:
$\frac{1}{2} |-5(5 - 7) + p(7 - 1) + 10(1 - 5)| = 0$
$|-5(-2) + p(6) + 10(-4)| = 0$
$|10 + 6p - 40| = 0$
$|6p - 30| = 0$
$6p = 30$
$p = 5$.

(D) For three points $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ to be collinear,the area of the triangle formed by them must be zero.
The area of the triangle is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$.
Substituting the given points $(k, 3), (2, k),$ and $(-k, 3)$:
$\frac{1}{2} |k(k - 3) + 2(3 - 3) + (-k)(3 - k)| = 0$
$|k(k - 3) + 0 - k(3 - k)| = 0$
$|k^2 - 3k - 3k + k^2| = 0$
$|2k^2 - 6k| = 0$
$2k(k - 3) = 0$
Thus,$k = 0$ or $k = 3$.

(A) The equations of the three sides are $L_1: x = 2$,$L_2: y = -1$,and $L_3: x + 2y = 4$.
$1$. Intersection of $L_1$ and $L_2$: $x = 2, y = -1 \implies A(2, -1)$.
$2$. Intersection of $L_1$ and $L_3$: $x = 2, 2 + 2y = 4 \implies 2y = 2 \implies y = 1 \implies B(2, 1)$.
$3$. Intersection of $L_2$ and $L_3$: $y = -1, x + 2(-1) = 4 \implies x - 2 = 4 \implies x = 6 \implies C(6, -1)$.
Let the circumcentre be $O(x, y)$. The distance $OA^2 = OB^2 = OC^2$.
$OA^2 = (x - 2)^2 + (y + 1)^2$
$OB^2 = (x - 2)^2 + (y - 1)^2$
$OC^2 = (x - 6)^2 + (y + 1)^2$
Equating $OA^2 = OB^2$: $(x - 2)^2 + (y + 1)^2 = (x - 2)^2 + (y - 1)^2 \implies (y + 1)^2 = (y - 1)^2 \implies y^2 + 2y + 1 = y^2 - 2y + 1 \implies 4y = 0 \implies y = 0$.
Equating $OA^2 = OC^2$ with $y = 0$: $(x - 2)^2 + (0 + 1)^2 = (x - 6)^2 + (0 + 1)^2 \implies (x - 2)^2 = (x - 6)^2 \implies x^2 - 4x + 4 = x^2 - 12x + 36 \implies 8x = 32 \implies x = 4$.
Thus,the circumcentre is $(4, 0)$.

(B) The lines $x = 0$ (y-axis),$y = 0$ (x-axis),and $3x + 4y = 12$ form a right-angled triangle with vertices at $A(0, 0)$,$B(0, 3)$,and $C(4, 0)$.
Let the vertices be $A(x_1, y_1) = (0, 0)$,$B(x_2, y_2) = (0, 3)$,and $C(x_3, y_3) = (4, 0)$.
The lengths of the sides opposite to these vertices are:
$a = BC = \sqrt{(4-0)^2 + (0-3)^2} = \sqrt{16 + 9} = 5$
$b = AC = \sqrt{(4-0)^2 + (0-0)^2} = 4$
$c = AB = \sqrt{(0-0)^2 + (3-0)^2} = 3$
The coordinates of the incentre $(I_x, I_y)$ are given by:
$I_x = \frac{ax_1 + bx_2 + cx_3}{a + b + c} = \frac{5(0) + 4(0) + 3(4)}{5 + 4 + 3} = \frac{12}{12} = 1$
$I_y = \frac{ay_1 + by_2 + cy_3}{a + b + c} = \frac{5(0) + 4(3) + 3(0)}{5 + 4 + 3} = \frac{12}{12} = 1$
Thus,the incentre is $(1, 1)$.

(A) Let the lines be $L_1: 4x - 7y + 10 = 0$,$L_2: x + y - 5 = 0$,and $L_3: 7x + 4y - 15 = 0$.
First,check the slopes of the lines:
Slope of $L_1$ $(m_1)$ = $4/7$.
Slope of $L_3$ $(m_3)$ = $-7/4$.
Since $m_1 \times m_3 = (4/7) \times (-7/4) = -1$,the lines $L_1$ and $L_3$ are perpendicular to each other.
Therefore,the triangle is a right-angled triangle with the right angle at the intersection of $L_1$ and $L_3$.
The orthocentre of a right-angled triangle is the vertex where the right angle is formed.
Solving $4x - 7y = -10$ and $7x + 4y = 15$:
Multiply the first by $4$ and the second by $7$: $16x - 28y = -40$ and $49x + 28y = 105$.
Adding these gives $65x = 65$,so $x = 1$.
Substituting $x = 1$ into $4(1) - 7y = -10$ gives $-7y = -14$,so $y = 2$.
The orthocentre is $(1, 2)$.

(B) Let the vertices be $A(at_1t_2, a(t_1 + t_2))$,$B(at_2t_3, a(t_2 + t_3))$,and $C(at_3t_1, a(t_3 + t_1))$.
The slope of $BC$ is $m_{BC} = \frac{a(t_2 + t_3) - a(t_3 + t_1)}{at_2t_3 - at_3t_1} = \frac{a(t_2 - t_1)}{at_3(t_2 - t_1)} = \frac{1}{t_3}$.
The slope of the altitude from $A$ to $BC$ is $-t_3$. The equation of this altitude is $y - a(t_1 + t_2) = -t_3(x - at_1t_2)$,which simplifies to $t_3x + y = a(t_1 + t_2 + t_1t_2t_3)$.
Similarly,the equation of the altitude from $B$ to $AC$ is $t_1x + y = a(t_2 + t_3 + t_1t_2t_3)$.
Subtracting the two equations: $(t_3 - t_1)x = a(t_1 - t_3)$,so $x = -a$.
Substituting $x = -a$ into the first altitude equation: $-at_3 + y = a(t_1 + t_2 + t_1t_2t_3)$,which gives $y = a(t_1 + t_2 + t_3 + t_1t_2t_3)$.
Thus,the orthocentre is $(-a, a(t_1 + t_2 + t_3 + t_1t_2t_3))$.

(A) Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$AB = \sqrt{(2a - 2a)^2 + (6a - 4a)^2} = \sqrt{0 + (2a)^2} = 2a$.
$BC = \sqrt{(2a + \sqrt{3}a - 2a)^2 + (5a - 6a)^2} = \sqrt{(\sqrt{3}a)^2 + (-a)^2} = \sqrt{3a^2 + a^2} = \sqrt{4a^2} = 2a$.
$CA = \sqrt{(2a + \sqrt{3}a - 2a)^2 + (5a - 4a)^2} = \sqrt{(\sqrt{3}a)^2 + a^2} = \sqrt{3a^2 + a^2} = \sqrt{4a^2} = 2a$.
Since $AB = BC = CA = 2a$,the triangle is an equilateral triangle.
Every equilateral triangle is an acute angled triangle,as all its angles are $60^\circ$.

(C) Let the given opposite vertices be $A(3, 4)$ and $C(1, -1)$. The midpoint of $AC$ is $M = \left( \frac{3+1}{2}, \frac{4-1}{2} \right) = \left( 2, \frac{3}{2} \right)$.
Let the other two vertices be $B$ and $D$. In a square,the diagonals are equal and bisect each other at right angles.
The vector $\vec{AC} = (1-3, -1-4) = (-2, -5)$.
The vector $\vec{BD}$ must be perpendicular to $\vec{AC}$ and have the same length. $A$ vector perpendicular to $(-2, -5)$ is $(5, -2)$ or $(-5, 2)$.
Since the diagonals bisect each other,the vertices $B$ and $D$ are given by $M \pm \frac{1}{2} \vec{BD}'$,where $\vec{BD}'$ is a vector of length equal to $AC$ perpendicular to $AC$.
The length $AC = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4+25} = \sqrt{29}$.
The vector $\vec{BD}'$ has length $\sqrt{29}$ and is perpendicular to $\vec{AC}$. Let $\vec{BD}' = (5, -2)$.
Then $B, D = M \pm \frac{1}{2} (5, -2) = \left( 2 \pm \frac{5}{2}, \frac{3}{2} \mp 1 \right)$.
For the plus sign: $\left( 2 + \frac{5}{2}, \frac{3}{2} - 1 \right) = \left( \frac{9}{2}, \frac{1}{2} \right)$.
For the minus sign: $\left( 2 - \frac{5}{2}, \frac{3}{2} + 1 \right) = \left( -\frac{1}{2}, \frac{5}{2} \right)$.
Thus,the coordinates are $\left( \frac{9}{2}, \frac{1}{2} \right)$ and $\left( -\frac{1}{2}, \frac{5}{2} \right)$,which matches option $C$.

(D) The vertex $O$ is the intersection of $4x + 5y = 0$ and $7x + 2y = 0,$ which is the origin $(0, 0).$
Since the diagonal $11x + 7y = 9$ does not pass through the origin $(0, 0),$ it cannot be the diagonal $OB.$ Thus,$11x + 7y = 9$ is the equation of the diagonal $AC.$
Let the sides be $OA: 4x + 5y = 0$ and $OC: 7x + 2y = 0.$
Solving $AC$ with $OA$: $4x + 5y = 0$ and $11x + 7y = 9.$ Multiplying the first by $7$ and the second by $5$: $28x + 35y = 0$ and $55x + 35y = 45.$ Subtracting gives $27x = 45 \implies x = \frac{5}{3}.$ Then $y = -\frac{4}{3}.$ So $A = \left(\frac{5}{3}, -\frac{4}{3}\right).$
Solving $AC$ with $OC$: $7x + 2y = 0$ and $11x + 7y = 9.$ Multiplying the first by $7$ and the second by $2$: $49x + 14y = 0$ and $22x + 14y = 18.$ Subtracting gives $27x = -18 \implies x = -\frac{2}{3}.$ Then $y = \frac{7}{3}.$ So $C = \left(-\frac{2}{3}, \frac{7}{3}\right).$
The midpoint $M$ of diagonal $AC$ is $\left(\frac{5/3 - 2/3}{2}, \frac{-4/3 + 7/3}{2}\right) = \left(\frac{1/3}{2}, \frac{3/3}{2}\right) = \left(\frac{1}{6}, \frac{1}{2}\right).$
The other diagonal $OB$ passes through the origin $(0, 0)$ and the midpoint $M\left(\frac{1}{6}, \frac{1}{2}\right).$
The slope of $OB$ is $m = \frac{1/2 - 0}{1/6 - 0} = \frac{1/2}{1/6} = 3.$
The equation of $OB$ is $y = 3x,$ or $3x - y = 0.$ Since this is not among the options,the correct answer is $D$ (None of these).

(A) The vertices of the quadrilateral are formed by the intersection of the lines:
$1$. $x=0$ and $y=0$ intersect at $B(0,0)$.
$2$. $x=0$ and $6x+y=3$ intersect at $A(0,3)$.
$3$. $y=0$ and $x+y=1$ intersect at $C(1,0)$.
$4$. $x+y=1$ and $6x+y=3$ intersect at $D(\frac{2}{5}, \frac{3}{5})$.
The diagonal passing through the origin $B(0,0)$ must also pass through the vertex $D(\frac{2}{5}, \frac{3}{5})$.
The equation of a line passing through $(0,0)$ and $(x_1, y_1)$ is $y = \frac{y_1}{x_1}x$.
Substituting $x_1 = \frac{2}{5}$ and $y_1 = \frac{3}{5}$:
$y = \frac{3/5}{2/5}x$
$y = \frac{3}{2}x$
$2y = 3x$
$3x - 2y = 0$.

(C) Let the given vertex be $A = (3, 4)$.
The equation of the first diagonal $d_1$ is $x + 2y = 1$,which has a slope $m_1 = -\frac{1}{2}$.
In a square,the diagonals are perpendicular to each other.
Therefore,the slope of the second diagonal $d_2$ is $m_2 = -\frac{1}{m_1} = -\frac{1}{-1/2} = 2$.
The second diagonal passes through the vertex $A(3, 4)$.
Using the point-slope form,the equation of the second diagonal is $y - 4 = 2(x - 3)$.
$y - 4 = 2x - 6$.
$2x - y - 2 = 0$,or $2x - y = 2$.

(B) The slope of the given side $x + y = 2$ is $m_1 = -1$,which corresponds to an angle of inclination $\theta = 135^\circ$.
Since the triangle is equilateral,the other two sides make an angle of $60^\circ$ with the base.
The slopes of the other two sides are given by $m = \tan(135^\circ \pm 60^\circ)$.
For the first case,$m = \tan(135^\circ - 60^\circ) = \tan(75^\circ) = 2 + \sqrt{3}$.
For the second case,$m = \tan(135^\circ + 60^\circ) = \tan(195^\circ) = \tan(15^\circ) = 2 - \sqrt{3}$.
Using the point-slope form $y - y_1 = m(x - x_1)$ with the vertex $(2, 3)$,the equations are $y - 3 = (2 \pm \sqrt{3})(x - 2)$.
Thus,one of the equations is $y - 3 = (2 - \sqrt{3})(x - 2)$.

(D) Let the vertices be $O(0, 0)$,$A(1, 0)$,$B(2, 2)$,and $C(1, 2)$.
The diagonals are $OB$ and $AC$.
The slope of diagonal $OB$ $(m_1)$ is $\frac{2 - 0}{2 - 0} = 1$.
The slope of diagonal $AC$ $(m_2)$ is $\frac{2 - 0}{1 - 1} = \frac{2}{0} = \infty$ (vertical line).
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by $\tan \theta = |\frac{m_2 - m_1}{1 + m_1 m_2}|$.
Since one line is vertical,the angle is $|90^\circ - \arctan(1)| = |90^\circ - 45^\circ| = 45^\circ$.
Thus,the angle is $\frac{\pi}{4}$.

(B) Let the vertices of the base be $A(2a, 0)$ and $B(0, a)$.
Since one side is $x = 2a$,the third vertex $C$ must lie on this line. Let $C = (2a, t)$.
For an isosceles triangle,the distance from the third vertex to the base vertices must be equal,or the vertex must lie on the perpendicular bisector.
Given $AC = BC$,we have $AC^2 = BC^2$.
$(2a - 2a)^2 + (t - 0)^2 = (2a - 0)^2 + (t - a)^2$
$t^2 = 4a^2 + t^2 - 2at + a^2$
$2at = 5a^2 \Rightarrow t = \frac{5a}{2}$.
Thus,the third vertex is $C(2a, \frac{5a}{2})$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area $= \frac{1}{2} |2a(0 - \frac{5a}{2}) + 2a(\frac{5a}{2} - 0) + 0(0 - 0)| = \frac{1}{2} |-5a^2 + 5a^2| = 0$ (This implies the points are collinear).
Re-evaluating: The side $x=2a$ connects $(2a, 0)$ and $(2a, t)$. The base is the segment connecting $(2a, 0)$ and $(0, a)$.
Length of base $AB = \sqrt{(2a-0)^2 + (0-a)^2} = \sqrt{4a^2 + a^2} = a\sqrt{5}$.
Height $h$ from $C(2a, t)$ to line $AB$ $(x + 2y - 2a = 0)$: $h = \frac{|2a + 2t - 2a|}{\sqrt{1^2 + 2^2}} = \frac{2t}{\sqrt{5}}$.
Since $AC = BC$,$C$ lies on the perpendicular bisector of $AB$. The slope of $AB$ is $m = \frac{a-0}{0-2a} = -1/2$. The perpendicular bisector has slope $2$ and passes through midpoint $(a, a/2)$.
Equation: $y - a/2 = 2(x - a) \Rightarrow y = 2x - 3a/2$.
Intersection with $x = 2a$: $y = 2(2a) - 3a/2 = 4a - 1.5a = 2.5a = \frac{5a}{2}$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (a\sqrt{5}) \times \frac{|2a + 2(5a/2) - 2a|}{\sqrt{5}} = \frac{1}{2} \times a\sqrt{5} \times \frac{5a}{\sqrt{5}} = \frac{5a^2}{2}$.

(D) Let the sides be $L_1: x - 3y = 0$,$L_2: 4x + 3y = 5$,and $L_3: 3x + y = 0$.
Check the slopes of the lines:
Slope of $L_1$ is $m_1 = 1/3$.
Slope of $L_3$ is $m_3 = -3$.
Since $m_1 \times m_3 = (1/3) \times (-3) = -1$,the lines $L_1$ and $L_3$ are perpendicular.
In a right-angled triangle,the orthocentre is the vertex where the right angle is formed.
The intersection of $L_1$ and $L_3$ is $(0, 0)$.
Thus,the orthocentre is $(0, 0)$.
Now,check if the line $3x - 4y = 0$ passes through $(0, 0)$:
$3(0) - 4(0) = 0 - 0 = 0$.
Since the point $(0, 0)$ satisfies the equation,the line passes through the orthocentre.

(D) The distance between the parallel lines $a_1x + b_1y + c_1 = 0$ and $a_1x + b_1y + d_1 = 0$ is $p_1 = \frac{|d_1 - c_1|}{\sqrt{a_1^2 + b_1^2}}$.
The distance between the parallel lines $a_2x + b_2y + c_2 = 0$ and $a_2x + b_2y + d_2 = 0$ is $p_2 = \frac{|d_2 - c_2|}{\sqrt{a_2^2 + b_2^2}}$.
The area of a parallelogram formed by two pairs of parallel lines is given by $\text{Area} = \frac{p_1 p_2}{\sin \theta}$,where $\theta$ is the angle between the lines.
The angle $\theta$ between the lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ satisfies $\sin \theta = \frac{|a_1b_2 - a_2b_1|}{\sqrt{a_1^2 + b_1^2} \sqrt{a_2^2 + b_2^2}}$.
Substituting $p_1$,$p_2$,and $\sin \theta$ into the area formula:
$\text{Area} = \frac{|d_1 - c_1|}{\sqrt{a_1^2 + b_1^2}} \times \frac{|d_2 - c_2|}{\sqrt{a_2^2 + b_2^2}} \times \frac{\sqrt{a_1^2 + b_1^2} \sqrt{a_2^2 + b_2^2}}{|a_1b_2 - a_2b_1|} = \frac{|(d_1 - c_1)(d_2 - c_2)|}{|a_1b_2 - a_2b_1|}$.

(A) The area of a parallelogram formed by two pairs of parallel lines $L_1: a_1x + b_1y + c_1 = 0, L_2: a_1x + b_1y + c_2 = 0$ and $M_1: a_2x + b_2y + d_1 = 0, M_2: a_2x + b_2y + d_2 = 0$ is given by $\text{Area} = \frac{|c_1 - c_2| |d_1 - d_2|}{|a_1b_2 - a_2b_1|}$.
Here,the lines are:
$x \cos \alpha + y \sin \alpha - p = 0$
$x \cos \alpha + y \sin \alpha - q = 0$
$x \cos \beta + y \sin \beta - r = 0$
$x \cos \beta + y \sin \beta - s = 0$
Distance between the first pair of parallel lines is $d_1 = |p - q|$.
Distance between the second pair of parallel lines is $d_2 = |r - s|$.
The angle $\theta$ between the lines is $\alpha - \beta$.
The denominator is $|(\cos \alpha)(\sin \beta) - (\cos \beta)(\sin \alpha)| = |\sin(\beta - \alpha)| = |\sin(\alpha - \beta)|$.
Thus,$\text{Area} = \frac{|p - q| |r - s|}{|\sin(\alpha - \beta)|} = |p - q| |r - s| |\csc(\alpha - \beta)|$.

(B) Given the equation of the line is $ax + by + c = 0$.
To find the $x$-intercept,set $y = 0$:
$ax + c = 0 \implies x = -\frac{c}{a}$.
To find the $y$-intercept,set $x = 0$:
$by + c = 0 \implies y = -\frac{c}{b}$.
The vertices of the triangle formed with the coordinate axes are $(0, 0)$,$(-\frac{c}{a}, 0)$,and $(0, -\frac{c}{b})$.
The area of the right-angled triangle is given by $\frac{1}{2} \times |\text{base}| \times |\text{height}|$.
Area $= \frac{1}{2} \times |-\frac{c}{a}| \times |-\frac{c}{b}|$.
Area $= \frac{1}{2} \left| \frac{c^2}{ab} \right| = \frac{c^2}{2|ab|}$.

(A) The vertices of the triangle are the intersection points of the given lines.
Solving the pairs of equations:
$1$) $x + y = 4$ and $3x + y = 4$: Subtracting gives $2x = 0 \implies x = 0, y = 4$. Vertex $A(0, 4)$.
$2$) $3x + y = 4$ and $x + 3y = 4$: Solving gives $x = 1, y = 1$. Vertex $B(1, 1)$.
$3$) $x + y = 4$ and $x + 3y = 4$: Subtracting gives $2y = 0 \implies y = 0, x = 4$. Vertex $C(4, 0)$.
Now,calculate the lengths of the sides:
$AB = \sqrt{(1-0)^2 + (1-4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{10}$
$BC = \sqrt{(4-1)^2 + (0-1)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{10}$
$AC = \sqrt{(4-0)^2 + (0-4)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2}$
Since $AB = BC$,the triangle is isosceles.