Problems related to triangle and quadrilateral Questions in English

(C) The equation of the hypotenuse $BC$ is $3x + 4y - 4 = 0$.
Let $A = (2, 2)$ be the vertex opposite to the hypotenuse.
The altitude $AD$ from $A$ to $BC$ is perpendicular to $BC$.
The slope of $BC$ is $m_{BC} = -\frac{3}{4}$.
Since $AD \perp BC$,the slope of $AD$ is $m_{AD} = \frac{4}{3}$.
In an isosceles right-angled triangle,the altitude from the vertex to the hypotenuse bisects the angle at the vertex. Thus,the sides $AB$ and $AC$ make an angle of $45^{\circ}$ with the altitude $AD$.
Let $m$ be the slope of side $AB$ or $AC$. Using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$:
$\tan 45^{\circ} = \left| \frac{m - \frac{4}{3}}{1 + m(\frac{4}{3})} \right|$ $\Rightarrow 1 = \left| \frac{3m - 4}{3 + 4m} \right|$.
This gives two cases:
$1 = \frac{3m - 4}{3 + 4m}$ $\Rightarrow 3 + 4m = 3m - 4$ $\Rightarrow m = -7$.
$-1 = \frac{3m - 4}{3 + 4m}$ $\Rightarrow -3 - 4m = 3m - 4$ $\Rightarrow 7m = 1$ $\Rightarrow m = \frac{1}{7}$.
Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(2, 2)$:
For $m = -7$: $y - 2 = -7(x - 2)$ $\Rightarrow y - 2 = -7x + 14$ $\Rightarrow 7x + y - 16 = 0$.
For $m = \frac{1}{7}$: $y - 2 = \frac{1}{7}(x - 2)$ $\Rightarrow 7y - 14 = x - 2$ $\Rightarrow x - 7y + 12 = 0$.
Comparing with the options,$7x + y - 16 = 0$ is present.

(B) Let the vertices be $A(1,3), B(5,0)$,and $C(-1,2)$.
For any point $(x,y)$ inside the triangle,the expression must maintain the same sign as it does for the vertices if the line does not pass through the triangle.
Testing the expression $3x + 2y$ for the vertices:
For $A(1,3): 3(1) + 2(3) = 3 + 6 = 9 > 0$.
For $B(5,0): 3(5) + 2(0) = 15 + 0 = 15 > 0$.
For $C(-1,2): 3(-1) + 2(2) = -3 + 4 = 1 > 0$.
Since the value is positive for all vertices,any point inside the triangle must satisfy $3x + 2y > 0$.
Therefore,option $B$ is correct.

(C) The vertices of the square are $O(0,0), A(1,0), B(1,1),$ and $C(0,1)$.
The diagonals are the line segments connecting opposite vertices,which are $OB$ and $AC$.
$1$. Equation of diagonal $OB$ passing through $(0,0)$ and $(1,1)$:
The slope $m = \frac{1-0}{1-0} = 1$.
Using the point-slope form $y - y_1 = m(x - x_1)$:
$y - 0 = 1(x - 0) \Rightarrow y = x$.
$2$. Equation of diagonal $AC$ passing through $(1,0)$ and $(0,1)$:
The slope $m = \frac{1-0}{0-1} = -1$.
Using the point-slope form $y - 0 = -1(x - 1)$:
$y = -x + 1 \Rightarrow x + y = 1$.
Thus,the equations of the diagonals are $y = x$ and $x + y = 1$.

(B) The sides of a square make an angle of $45^{\circ}$ with the diagonal. The slope of the given diagonal $8x - 15y = 0$ is $m_1 = \frac{8}{15}$.
Let the slope of the sides passing through $(1, 2)$ be $m$. Using the formula $\tan(45^{\circ}) = |\frac{m - m_1}{1 + m \cdot m_1}|$,we get:
$1 = |\frac{m - 8/15}{1 + 8m/15}|$
$1 = |\frac{15m - 8}{15 + 8m}|$
This gives two cases:
Case $1$: $15m - 8 = 15 + 8m$ $\Rightarrow 7m = 23$ $\Rightarrow m = \frac{23}{7}$.
The equation of the side is $y - 2 = \frac{23}{7}(x - 1)$ $\Rightarrow 7y - 14 = 23x - 23$ $\Rightarrow 23x - 7y - 9 = 0$.
Case $2$: $15m - 8 = -(15 + 8m)$ $\Rightarrow 15m - 8 = -15 - 8m$ $\Rightarrow 23m = -7$ $\Rightarrow m = -\frac{7}{23}$.
The equation of the side is $y - 2 = -\frac{7}{23}(x - 1)$ $\Rightarrow 23y - 46 = -7x + 7$ $\Rightarrow 7x + 23y - 53 = 0$.

(B) The given lines are $L_1: 4x+y-1=0$ and $L_2: 3x-4y+1=0$. The slope of $L_1$ is $m_1 = -4$ and the slope of $L_2$ is $m_2 = \frac{3}{4}$.
Point $A$ is $(2, -7)$. Line $AB$ is $L_1$. Point $B$ is the intersection of $L_1$ and $L_2$.
Since $AB=AC$ and $A$ is common,the angle between $AB$ and $BC$ must be equal to the angle between $AC$ and $BC$. Let the slope of $AC$ be $m$. The angle $\theta$ between $AB$ and $BC$ is given by $\tan \theta = |\frac{m_1-m_2}{1+m_1m_2}| = |\frac{-4 - 3/4}{1 + (-4)(3/4)}| = |\frac{-19/4}{1-3}| = |\frac{-19/4}{-2}| = \frac{19}{8}$.
Since $BC$ is the base of the isosceles triangle $ABC$,the line $BC$ makes the same angle with $AB$ and $AC$. Thus,the angle between $AC$ and $BC$ is also $\theta = \tan^{-1}(\frac{19}{8})$.
Using $\tan \theta = |\frac{m-m_2}{1+mm_2}|$,we have $\frac{19}{8} = |\frac{m-3/4}{1+m(3/4)}| = |\frac{4m-3}{4+3m}|$.
This gives two cases: $\frac{4m-3}{4+3m} = \frac{19}{8}$ or $\frac{4m-3}{4+3m} = -\frac{19}{8}$.
Case $1$: $32m - 24 = 76 + 57m$ $\Rightarrow -25m = 100$ $\Rightarrow m = -4$ (This is line $AB$).
Case $2$: $32m - 24 = -76 - 57m$ $\Rightarrow 89m = -52$ $\Rightarrow m = -\frac{52}{89}$.
The equation of line $AC$ is $y - (-7) = -\frac{52}{89}(x - 2)$ $\Rightarrow 89(y+7) = -52(x-2)$ $\Rightarrow 89y + 623 = -52x + 104$ $\Rightarrow 52x + 89y + 519 = 0$.

(B) The perpendicular bisector of $AB$ is $2x+3y+1=0$. The slope of this line is $-2/3$. Thus,the slope of $AB$ is $3/2$.
The equation of $AB$ is $y-2 = \frac{3}{2}(x-3)$,which simplifies to $3x-2y-5=0$.
The intersection of $AB$ $(3x-2y-5=0)$ and its perpendicular bisector $(2x+3y+1=0)$ gives the midpoint $D$ of $AB$. Solving these,we get $D=(1,-1)$.
Since $D$ is the midpoint of $AB$,$\frac{3+x_B}{2}=1$ and $\frac{2+y_B}{2}=-1$,so $B=(-1,-4)$.
The perpendicular bisector of $AC$ is $x+2y-2=0$. The slope of this line is $-1/2$. Thus,the slope of $AC$ is $2$.
The equation of $AC$ is $y-2 = 2(x-3)$,which simplifies to $2x-y-4=0$.
The intersection of $AC$ $(2x-y-4=0)$ and its perpendicular bisector $(x+2y-2=0)$ gives the midpoint $E$ of $AC$. Solving these,we get $E=(2,0)$.
Since $E$ is the midpoint of $AC$,$\frac{3+x_C}{2}=2$ and $\frac{2+y_C}{2}=0$,so $C=(1,-2)$.
The equation of side $BC$ passing through $B(-1,-4)$ and $C(1,-2)$ is $y-(-4) = \frac{-2-(-4)}{1-(-1)}(x-(-1))$.
$y+4 = \frac{2}{2}(x+1) \implies y+4 = x+1 \implies x-y-3=0$.

(B) Let the mid-points be $P(2,1)$ on $BC$,$Q(-1,-2)$ on $CA$,and $R(3,3)$ on $AB$.
Since $RQ$ is parallel to $BC$ and $RQ = \frac{1}{2} BC$,the side $BC$ is parallel to the line segment $RQ$.
The slope of $RQ$ is $m = \frac{3 - (-2)}{3 - (-1)} = \frac{5}{4}$.
Since $BC$ is parallel to $RQ$,the slope of $BC$ is also $m = \frac{5}{4}$.
The side $BC$ passes through the point $P(2,1)$.
Using the point-slope form,the equation of $BC$ is:
$y - 1 = \frac{5}{4}(x - 2)$
$4(y - 1) = 5(x - 2)$
$4y - 4 = 5x - 10$
$5x - 4y = 6$

(B) Let $C = (h, k)$. Since $C$ lies on the angle bisector $7x - 4y - 1 = 0$,we have $7h - 4k - 1 = 0$,or $h = \frac{4k + 1}{7}$.
Since $M$ is the midpoint of $AC$,$M = \left(\frac{-3 + h}{2}, \frac{1 + k}{2}\right) = \left(\frac{-3 + \frac{4k+1}{7}}{2}, \frac{1+k}{2}\right) = \left(\frac{4k - 20}{14}, \frac{1+k}{2}\right) = \left(\frac{2k - 10}{7}, \frac{1+k}{2}\right)$.
Since $M$ lies on the median $BM$ $(2x + y - 3 = 0)$,we have $2\left(\frac{2k - 10}{7}\right) + \frac{1+k}{2} - 3 = 0$.
Multiplying by $14$,we get $4(2k - 10) + 7(1+k) - 42 = 0$ $\Rightarrow 8k - 40 + 7 + 7k - 42 = 0$ $\Rightarrow 15k = 75$ $\Rightarrow k = 5$.
Then $h = \frac{4(5) + 1}{7} = 3$. So,$C = (3, 5)$.
The slope of $AC$ is $m_{AC} = \frac{5 - 1}{3 - (-3)} = \frac{4}{6} = \frac{2}{3}$.
The slope of the angle bisector $CN$ is $m_{bisector} = \frac{7}{4}$.
Let $m$ be the slope of $BC$. Using the angle bisector property $\tan(\angle BCN) = \tan(\angle NCA)$:
$\left|\frac{m - 7/4}{1 + m(7/4)}\right| = \left|\frac{7/4 - 2/3}{1 + (7/4)(2/3)}\right| = \left|\frac{21 - 8}{12 + 14}\right| = \frac{13}{26} = \frac{1}{2}$.
Taking the positive case: $\frac{4m - 7}{4 + 7m} = \frac{1}{2}$ $\Rightarrow 8m - 14 = 4 + 7m$ $\Rightarrow m = 18$.
The equation of $BC$ passing through $C(3, 5)$ with slope $18$ is $y - 5 = 18(x - 3)$ $\Rightarrow y - 5 = 18x - 54$ $\Rightarrow 18x - y = 49$.

(B) Since $\triangle PQR$ is an isosceles right-angled triangle with the right angle at $P$,we have $\angle PQR = \angle PRQ = 45^{\circ}$.
Let $m$ be the slope of the line $QR$. Given $2x + y = 3$,we have $y = -2x + 3$,so $m = -2$.
Let $m_1$ be the slope of the line $PQ$. The angle between $PQ$ and $QR$ is $45^{\circ}$.
Using the formula $\tan \theta = |\frac{m_1 - m}{1 + m_1 m}|$,we have:
$\tan 45^{\circ} = |\frac{m_1 - (-2)}{1 + m_1(-2)}| = |\frac{m_1 + 2}{1 - 2m_1}|$
$1 = |\frac{m_1 + 2}{1 - 2m_1}|$
This gives two cases:
Case $1$: $\frac{m_1 + 2}{1 - 2m_1} = 1$ $\Rightarrow m_1 + 2 = 1 - 2m_1$ $\Rightarrow 3m_1 = -1$ $\Rightarrow m_1 = -\frac{1}{3}$.
Case $2$: $\frac{m_1 + 2}{1 - 2m_1} = -1$ $\Rightarrow m_1 + 2 = -1 + 2m_1$ $\Rightarrow m_1 = 3$.
Since $PQ$ and $PR$ are perpendicular and pass through $P(2, 1)$,their equations are:
For $m_1 = -\frac{1}{3}$: $y - 1 = -\frac{1}{3}(x - 2)$ $\Rightarrow 3y - 3 = -x + 2$ $\Rightarrow x + 3y - 5 = 0$.
For $m_2 = 3$: $y - 1 = 3(x - 2)$ $\Rightarrow y - 1 = 3x - 6$ $\Rightarrow 3x - y - 5 = 0$.
Comparing with the options,$3x - y - 5 = 0$ is the correct equation.

(D) The area of a parallelogram formed by the lines $a_1x+b_1y+c_1=0$,$a_1x+b_1y+c_2=0$,$a_2x+b_2y+d_1=0$,and $a_2x+b_2y+d_2=0$ is given by the formula: $\text{Area} = \left| \frac{(c_1-c_2)(d_1-d_2)}{a_1b_2-a_2b_1} \right|$.
Here,the lines are $ax+by+2=0$,$ax+by+5=0$,$cx+dy+3=0$,and $cx+dy+7=0$.
Comparing with the formula,we have $c_1=2, c_2=5, d_1=3, d_2=7, a_1=a, b_1=b, a_2=c, b_2=d$.
Substituting these values into the formula:
$\text{Area} = \left| \frac{(2-5)(3-7)}{ad-bc} \right|$
$\text{Area} = \left| \frac{(-3)(-4)}{ad-bc} \right|$
$\text{Area} = \frac{12}{|ad-bc|}$ sq. units.

(A) Let $A = (1, -2)$. The perpendicular bisector of $AB$ is $L_1: x-y+5=0$. The slope of $L_1$ is $1$,so the slope of $AB$ is $-1$. The equation of $AB$ is $y - (-2) = -1(x - 1) \Rightarrow x+y+1=0$.
Intersection of $AB$ and $L_1$: $x - (-x-1) + 5 = 0$ $\Rightarrow 2x = -6$ $\Rightarrow x = -3, y = 2$.
Since $E$ is the midpoint of $AB$,$\frac{x_B+1}{2} = -3 \Rightarrow x_B = -7$ and $\frac{y_B-2}{2} = 2 \Rightarrow y_B = 6$. Thus $B = (-7, 6)$.
The perpendicular bisector of $AC$ is $L_2: x+2y=0$. The slope of $L_2$ is $-1/2$,so the slope of $AC$ is $2$. The equation of $AC$ is $y - (-2) = 2(x - 1) \Rightarrow 2x-y-4=0$.
Intersection of $AC$ and $L_2$: $x + 2(2x-4) = 0$ $\Rightarrow 5x = 8$ $\Rightarrow x = 8/5, y = -4/5$.
Since $F$ is the midpoint of $AC$,$\frac{x_C+1}{2} = 8/5 \Rightarrow x_C = 11/5$ and $\frac{y_C-2}{2} = -4/5 \Rightarrow y_C = 2/5$. Thus $C = (11/5, 2/5)$.
The equation of line $BC$ passing through $(-7, 6)$ and $(11/5, 2/5)$ is:
$y - 6 = \frac{2/5 - 6}{11/5 - (-7)} (x - (-7))
$ $\Rightarrow y - 6 = \frac{-28/5}{46/5} (x + 7)
$ $\Rightarrow y - 6 = -\frac{14}{23} (x + 7)
$ $\Rightarrow 23y - 138 = -14x - 98
$ $\Rightarrow 14x + 23y - 40 = 0$.

(D) Let $ABCD$ be a square where $A \equiv (1,3)$ and $C \equiv (7,5)$.
Diagonal $AC$ has slope $m_{AC} = \frac{5-3}{7-1} = \frac{2}{6} = \frac{1}{3}$.
Let the slope of the side $AB$ be $m$.
Since the diagonal of a square makes an angle of $45^{\circ}$ with the sides,we have $\left| \frac{m - 1/3}{1 + m(1/3)} \right| = \tan 45^{\circ} = 1$.
$\left| \frac{3m-1}{3+m} \right| = 1$.
Case $1$: $\frac{3m-1}{3+m} = 1$ $\Rightarrow 3m-1 = 3+m$ $\Rightarrow 2m = 4$ $\Rightarrow m = 2$.
The equation of the side passing through $A(1,3)$ with slope $m=2$ is $y-3 = 2(x-1)$ $\Rightarrow y-3 = 2x-2$ $\Rightarrow 2x-y+1 = 0$.
Case $2$: $\frac{3m-1}{3+m} = -1$ $\Rightarrow 3m-1 = -3-m$ $\Rightarrow 4m = -2$ $\Rightarrow m = -1/2$.
The equation of the side passing through $A(1,3)$ with slope $m=-1/2$ is $y-3 = -1/2(x-1)$ $\Rightarrow 2y-6 = -x+1$ $\Rightarrow x+2y-7 = 0$.
Comparing with the options,$2x-y+1=0$ is present.

(C) The given line is $4x - 2y = 1$,which can be written as $2y = 4x - 1$ or $y = 2x - 1/2$. The slope of this line is $m_1 = 2$.
Since line $L$ is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$. Thus,$2 \times m_2 = -1$,so $m_2 = -1/2$.
The equation of line $L$ with slope $-1/2$ can be written as $x + 2y + \lambda = 0$.
The intercepts of this line on the coordinate axes are found by setting $x=0$ and $y=0$:
For $x=0$,$2y = -\lambda \implies y = -\lambda/2$.
For $y=0$,$x = -\lambda$.
The area of the triangle formed with the coordinate axes is $\frac{1}{2} \times |\text{base}| \times |\text{height}| = \frac{1}{2} |-\lambda| \times |-\lambda/2| = 4$.
$\frac{\lambda^2}{4} = 4 \implies \lambda^2 = 16 \implies \lambda = \pm 4$.
Substituting $\lambda = 4$ into $x + 2y + \lambda = 0$ gives $x + 2y + 4 = 0$,which is equivalent to $2x + 4y + 8 = 0$.
Substituting $\lambda = -4$ gives $x + 2y - 4 = 0$,which is equivalent to $2x + 4y - 8 = 0$.

(C) The equations of the two sides of the rhombus are $L_1: x-y+1=0$ and $L_2: 7x-y-5=0$.
Solving these equations,we find the vertex $V_1$ where these sides intersect:
$x-y = -1$ and $7x-y = 5$.
Subtracting the first from the second gives $6x = 6$,so $x=1$.
Substituting $x=1$ into $x-y+1=0$ gives $y=2$. Thus,$V_1 = (1,2)$.
The diagonals of a rhombus bisect each other at the intersection point $P(-1,-2)$.
Since $P$ is the midpoint of the diagonal connecting $V_1(1,2)$ and its opposite vertex $V_3(x_3, y_3)$,we have:
$-1 = \frac{1+x_3}{2} \Rightarrow x_3 = -3$ and $-2 = \frac{2+y_3}{2} \Rightarrow y_3 = -6$.
So,$V_3 = (-3,-6)$.
The other diagonal passes through $P(-1,-2)$ and is perpendicular to the diagonal $V_1V_3$.
The slope of $V_1V_3$ is $m = \frac{-6-2}{-3-1} = \frac{-8}{-4} = 2$.
The slope of the other diagonal is $m' = -\frac{1}{2}$.
The equation of the other diagonal is $y - (-2) = -\frac{1}{2}(x - (-1))$,which simplifies to $2y + 4 = -x - 1$,or $x + 2y + 5 = 0$.
The other two vertices lie on this diagonal. Checking the options,for $C \left(\frac{1}{3}, -\frac{8}{3}\right)$:
$\frac{1}{3} + 2\left(-\frac{8}{3}\right) + 5 = \frac{1-16+15}{3} = 0$.
Thus,$\left(\frac{1}{3}, -\frac{8}{3}\right)$ is a vertex.

(A) The triangle is formed by the lines $x=0$,$y=0$,and $x+y=10$. The vertices are $O(0,0)$,$A(10,0)$,and $B(0,10)$.
We need to find the number of points $(x, y)$ such that $x, y \in \mathbb{N}$ (natural numbers) and $x+y < 10$.
If $x=1$,then $1+y < 10 \implies y < 9$. Since $y$ is a natural number,$y \in \{1, 2, 3, 4, 5, 6, 7, 8\}$. There are $8$ such points.
If $x=2$,then $2+y < 10 \implies y < 8$. Thus $y \in \{1, 2, 3, 4, 5, 6, 7\}$. There are $7$ such points.
Continuing this pattern,for a given $x$,the number of natural number values for $y$ is $9-x$.
Since $x$ must be a natural number,$x$ can range from $1$ to $8$ (because if $x=9$,$y < 1$,which is impossible for natural numbers).
The total number of points is the sum: $8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = \frac{8 \times 9}{2} = 36$.

(B) Let the lines be $L_1: 2x + 3y - 1 = 0$,$L_2: x + 2y + 1 = 0$,and $L_3: ax + by - 1 = 0$. The orthocenter is at $(0, 0)$.
Since the orthocenter is the intersection of altitudes,the altitude from vertex $B$ (intersection of $L_1$ and $L_2$) passes through $(0, 0)$ and is perpendicular to $L_3$. The family of lines through $B$ is $(2x + 3y - 1) + \lambda(x + 2y + 1) = 0$. Since it passes through $(0, 0)$,we have $-1 + \lambda = 0 \Rightarrow \lambda = 1$. The altitude is $3x + 5y = 0$,with slope $m = -\frac{3}{5}$. Since this is perpendicular to $L_3$ (slope $-\frac{a}{b}$),we have $(-\frac{a}{b}) \times (-\frac{3}{5}) = -1 \Rightarrow 3a = -5b$.
Similarly,the altitude from vertex $A$ (intersection of $L_1$ and $L_3$) passes through $(0, 0)$ and is perpendicular to $L_2$ (slope $-\frac{1}{2}$). The family of lines through $A$ is $(2x + 3y - 1) + \mu(ax + by - 1) = 0$. Since it passes through $(0, 0)$,we have $-1 - \mu = 0 \Rightarrow \mu = -1$. The altitude is $(2-a)x + (3-b)y = 0$,with slope $-\frac{2-a}{3-b}$. Since this is perpendicular to $L_2$,we have $(-\frac{2-a}{3-b}) \times (-\frac{1}{2}) = -1$ $\Rightarrow 2-a = -2(3-b)$ $\Rightarrow a + 2b = 8$.
Solving $3a = -5b$ and $a + 2b = 8$,we get $a = -40$ and $b = 24$. Wait,checking the calculation: $a = -40, b = 24 \Rightarrow 3(-40) = -120, -5(24) = -120$ (Correct). $a + 2b = -40 + 48 = 8$ (Correct). Thus,$\frac{1}{a} + \frac{1}{b} = -\frac{1}{40} + \frac{1}{24} = \frac{-3 + 5}{120} = \frac{2}{120} = \frac{1}{60}$.

(A) The condition $xy > 0$ implies that the point $A(x, y)$ must lie in either the first quadrant $(x > 0, y > 0)$ or the third quadrant $(x < 0, y < 0)$.
The condition $x + y < 1$ represents the region below the line $x + y = 1$.
In the first quadrant,the region satisfying $x > 0, y > 0$ and $x + y < 1$ is the interior of the triangle formed by the origin $O(0, 0)$,$(1, 0)$,and $(0, 1)$.
The triangle $\triangle OMN$ has vertices $O(0, 0)$,$M(1, 2)$,and $N(0, 1)$.
By observing the region defined by $xy > 0$ and $x + y < 1$,it is clear that the point $A$ cannot lie inside $\triangle OMN$ because the interior of $\triangle OMN$ lies above the line $x + y = 1$ for most of its area,and the conditions given restrict $A$ to a region that does not overlap with the interior of $\triangle OMN$.

(A) The triangle is formed by the intersection of the lines $L_1: 3x+y+2=0$,$L_2: 2x-3y+5=0$,and $L_3: x+4y-14=0$.
To find the range of $\beta$ for the point $(0, \beta)$ to lie inside the triangle,we find the $y$-intercepts of these lines on the $y$-axis (where $x=0$):
For $L_1: 3(0)+y+2=0 \implies y = -2$.
For $L_2: 2(0)-3y+5=0 \implies y = 5/3$.
For $L_3: 0+4y-14=0 \implies y = 14/4 = 7/2$.
By observing the region bounded by these lines,the point $(0, \beta)$ lies within the triangle when $\beta$ is between the two $y$-intercepts that bound the vertical segment of the triangle on the $y$-axis.
From the graph,the $y$-intercepts are $-2$,$5/3$,and $7/2$. The segment on the $y$-axis inside the triangle lies between $y = 5/3$ and $y = 7/2$.
Thus,the set of values for $\beta$ is $\left[\frac{5}{3}, \frac{7}{2}\right]$.

(A) The equation of the side is $x+y-2=0$.
The vertex is $V(2,-1)$.
The perpendicular distance $h$ from the vertex $(2,-1)$ to the line $x+y-2=0$ is the height of the equilateral triangle.
$h = \frac{|2 + (-1) - 2|}{\sqrt{1^2 + 1^2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
For an equilateral triangle with side length $a$,the height $h = \frac{\sqrt{3}}{2}a$.
Therefore,$\frac{\sqrt{3}}{2}a = \frac{1}{\sqrt{2}}$.
$a = \frac{2}{\sqrt{3} \times \sqrt{2}} = \frac{\sqrt{2}}{\sqrt{3}}$.

(A) Given that $\triangle ABC$ is an isosceles triangle with base $BC$ where $B \equiv (3, 2)$ and $C \equiv (2, 3)$.
Note that the points $B(3, 2)$ and $C(2, 3)$ are symmetric with respect to the line $y = x$.
Since the triangle is isosceles with base $BC$,the vertex $A$ must lie on the perpendicular bisector of $BC$. The perpendicular bisector of the segment joining $(3, 2)$ and $(2, 3)$ is the line $y = x$.
Because the entire configuration is symmetric about the line $y = x$,the line $AC$ is the reflection of the line $AB$ about the line $y = x$.
To find the reflection of the line $3y = 2x$ about $y = x$,we swap $x$ and $y$ in the equation.
Replacing $x$ with $y$ and $y$ with $x$,we get $3x = 2y$,which is $2y = 3x$.
Thus,the equation of the line $AC$ is $2y = 3x$.

(C) Let the lines be $L_1: 2x + 3y - 1 = 0$,$L_2: x + 2y - 1 = 0$,and $L_3: ax + by - 1 = 0$. The origin $(0, 0)$ is the orthocentre.
The altitude from the intersection of $L_1$ and $L_2$ to $L_3$ must pass through the origin.
The intersection of $L_1$ and $L_2$ is found by solving $2x + 3y = 1$ and $x + 2y = 1$,which gives $x = -1, y = 1$.
The line passing through $(-1, 1)$ and $(0, 0)$ is $y = -x$,or $x + y = 0$.
Since this line is perpendicular to $L_3: ax + by - 1 = 0$,the slope of $L_3$ is $-a/b$.
The slope of $x + y = 0$ is $-1$. Thus,$(-a/b) \times (-1) = -1$,which implies $a/b = -1$,or $a = -b$.
Similarly,the altitude from the intersection of $L_2$ and $L_3$ to $L_1$ passes through $(0, 0)$.
The intersection of $L_2$ and $L_3$ is $(x, y)$ such that $x + 2y = 1$ and $ax + by = 1$.
Using the property that the orthocentre is $(0, 0)$,we find $a = -8$ and $b = 8$.

(D) Since $BD=2 \sqrt{2}$,we have $OB=OD=\sqrt{2}$.
Given that $(3,4)$ is the midpoint of $AC$,we have $C=(5,6)$.
Thus,$OA=OC=\sqrt{(3-1)^2+(4-2)^2}=\sqrt{4+4}=2 \sqrt{2}$.
In $\triangle AOB$,$OA^2+OB^2=AB^2$,so $AB^2=(2\sqrt{2})^2+(\sqrt{2})^2=8+2=10$,which means $AB=\sqrt{10}$.
Since $O(3,4)$ is the midpoint of $BD$,we have $\alpha+\gamma=6$ and $\beta+\delta=8$.
Also,$OB^2=(\alpha-3)^2+(\beta-4)^2=2$ and $OD^2=(\gamma-3)^2+(\delta-4)^2=2$.
Since $ABCD$ is a rhombus,$AB=BC=CD=DA=\sqrt{10}$.
Using the distance formula for $CD^2=10$,we get $(\gamma-5)^2+(\delta-6)^2=10$.
Solving the system of equations for $\alpha, \beta, \gamma, \delta$ with the condition $\alpha < \delta < \gamma < \beta$,we find that $\beta-\delta=-2\alpha+6$.
Finally,$\beta+\gamma-\delta=(\beta-\delta)+\gamma=(-2\alpha+6)+(6-\alpha)=-3\alpha+12$.

(C) The slope of the side $AB$ is $m_{AB} = \frac{8-7}{-7-4} = \frac{1}{-11} = -\frac{1}{11}$.
Since the perpendicular bisector is perpendicular to $AB$,its slope $m$ is given by $m \times m_{AB} = -1$,so $m = 11$.
The midpoint $D$ of side $AB$ is $\left(\frac{4+(-7)}{2}, \frac{7+8}{2}\right) = \left(-\frac{3}{2}, \frac{15}{2}\right)$.
The equation of the line passing through $D\left(-\frac{3}{2}, \frac{15}{2}\right)$ with slope $m=11$ is:
$y - \frac{15}{2} = 11\left(x - (-\frac{3}{2}\right))$
$y - \frac{15}{2} = 11\left(x + \frac{3}{2}\right)$
$y - \frac{15}{2} = 11x + \frac{33}{2}$
$11x - y + \frac{33}{2} + \frac{15}{2} = 0$
$11x - y + \frac{48}{2} = 0$
$11x - y + 24 = 0$.

(B) The height $h$ of the equilateral triangle is the perpendicular distance from the vertex $(2,1)$ to the line $x+y-2=0$.
$h = \frac{|2+1-2|}{\sqrt{1^2+1^2}} = \frac{1}{\sqrt{2}}$.
Since $h = \frac{\sqrt{3}}{2} a$,we have $a = \frac{2h}{\sqrt{3}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}$.
Thus,$a \sqrt{2} = \sqrt{\frac{2}{3}} \times \sqrt{2} = \frac{2}{\sqrt{3}}$.
The slope of the base is $m = -1$. Let the slopes of the other two sides be $m_1$ and $m_2$.
The angle between the base and the sides is $60^\circ$.
Using $\tan 60^\circ = |\frac{m_1 - (-1)}{1 + m_1(-1)}| = |\frac{m_1+1}{1-m_1}| = \sqrt{3}$.
Solving $\frac{m_1+1}{1-m_1} = \sqrt{3}$ gives $m_1 = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}$.
Solving $\frac{m_1+1}{1-m_1} = -\sqrt{3}$ gives $m_2 = \frac{\sqrt{3}+1}{\sqrt{3}-1} = 2+\sqrt{3}$.
Then $|m_1-m_2| = |(2-\sqrt{3}) - (2+\sqrt{3})| = |-2\sqrt{3}| = 2\sqrt{3}$.
Finally,$|m_1-m_2| + a\sqrt{2} = 2\sqrt{3} + \frac{2}{\sqrt{3}} = \frac{6+2}{\sqrt{3}} = \frac{8}{\sqrt{3}}$.

(B) Given sides are $a = 4 \text{ cm}$,$b = 5 \text{ cm}$,and $c = 7 \text{ cm}$.
First,calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{4 + 5 + 7}{2} = \frac{16}{2} = 8 \text{ cm}$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$
$= \sqrt{8(8 - 4)(8 - 5)(8 - 7)}$
$= \sqrt{8 \times 4 \times 3 \times 1}$
$= \sqrt{96}$
$= \sqrt{16 \times 6}$
$= 4 \sqrt{6} \text{ cm}^2$.

(B) Given that the perimeter $2s = 16 \text{ cm}$,so the semi-perimeter $s = 8 \text{ cm}$.
Let the sides be $a, b, c$. Given $a = 6 \text{ cm}$ and area $\Delta = 12 \text{ cm}^2$.
Using Heron's formula: $\Delta^2 = s(s-a)(s-b)(s-c)$.
$144 = 8(8-6)(8-b)(8-c)$.
$144 = 8(2)(8-b)(8-c)$.
$144 = 16(8-b)(8-c) \Rightarrow 9 = (8-b)(8-c)$.
Since $a+b+c = 16$ and $a=6$,we have $b+c = 10$,so $c = 10-b$.
Substituting $c$ into the equation: $9 = (8-b)(8-(10-b))$.
$9 = (8-b)(b-2)$.
$9 = 8b - 16 - b^2 + 2b$.
$b^2 - 10b + 25 = 0$.
$(b-5)^2 = 0 \Rightarrow b = 5$.
Since $b=5$,then $c = 10-5 = 5$.
As two sides are equal $(b=c=5)$,the triangle is isosceles.

(B) The region is bounded by the lines $x=4$,$y=-4$,and $y=x$.
To find the vertices of the triangle,we find the intersection points of these lines:
$1$. Intersection of $x=4$ and $y=x$ is $B(4, 4)$.
$2$. Intersection of $y=-4$ and $y=x$ is $A(-4, -4)$.
$3$. Intersection of $x=4$ and $y=-4$ is $C(4, -4)$.
The region is a right-angled triangle $ABC$ with vertices $A(-4, -4)$,$B(4, 4)$,and $C(4, -4)$.
The length of the base $AC$ is the distance between $(-4, -4)$ and $(4, -4)$,which is $|4 - (-4)| = 8$ units.
The length of the height $BC$ is the distance between $(4, -4)$ and $(4, 4)$,which is $|4 - (-4)| = 8$ units.
The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Area $= \frac{1}{2} \times 8 \times 8 = 32$ sq. units.

(C) In a right-angled triangle,let the sides be $a = 2 \sqrt{2}$,$b = 5$,and the third side be $p$.
Case $1$: If $p$ is the hypotenuse,then by the Pythagorean theorem,$p^2 = a^2 + b^2$.
$p^2 = (2 \sqrt{2})^2 + 5^2 = 8 + 25 = 33$.
So,$p = \sqrt{33}$.
Case $2$: If $5$ is the hypotenuse,then $p^2 + a^2 = 5^2$.
$p^2 + 8 = 25 \Rightarrow p^2 = 17$.
So,$p = \sqrt{17}$.
Since $\sqrt{17}$ is one of the given options,option $(c)$ is correct.

(C) Given that $\angle CAB = \angle CBA$,the triangle $\triangle ABC$ is an isosceles triangle with $AC = BC$.
Squaring both sides,we get $AC^2 = BC^2$.
$(h-1)^2 + (k-3)^2 = (h-2)^2 + (k-5)^2$
$h^2 - 2h + 1 + k^2 - 6k + 9 = h^2 - 4h + 4 + k^2 - 10k + 25$
$-2h - 6k + 10 = -4h - 10k + 29$
$2h + 4k = 19 \implies k = \frac{19-2h}{4} \quad (1)$
Since $AC \perp BC$,the product of their slopes is $-1$:
$m_{AC} \times m_{BC} = -1$
$\frac{k-3}{h-1} \times \frac{k-5}{h-2} = -1$
$(k-3)(k-5) = -(h-1)(h-2)$
$k^2 - 8k + 15 = -(h^2 - 3h + 2) = -h^2 + 3h - 2$
$h^2 + k^2 - 3h - 8k + 17 = 0 \quad (2)$
Substituting $(1)$ into $(2)$:
$h^2 + (\frac{19-2h}{4})^2 - 3h - 8(\frac{19-2h}{4}) + 17 = 0$
$16h^2 + (361 - 76h + 4h^2) - 48h - 32(19-2h) + 272 = 0$
$20h^2 - 124h + 361 - 608 + 64h + 272 = 0$
$20h^2 - 60h + 25 = 0$
$4h^2 - 12h + 5 = 0$
$(2h-1)(2h-5) = 0$
$h = \frac{1}{2}$ or $h = \frac{5}{2}$.

(B) Let $O(1, 1)$ be the circumcenter. The perpendicular bisector of $AC$ passes through $O(1, 1)$ and the midpoint $P(a, \frac{b+3}{2})$. The slope of $AC$ is $\frac{b-3}{a-a}$,which is undefined (vertical line). Thus,the perpendicular bisector is horizontal: $y = \frac{b+3}{2} = 1$ $\Rightarrow b+3=2$ $\Rightarrow b=-1$.
Alternatively,from the image,the slope of $OP$ is $\frac{\frac{b+3}{2}-1}{a-1} = -\frac{a-a}{b-3} = 0$. This implies $\frac{b+3}{2} = 1$,so $b=-1$.
Using $OQ \perp AB$,where $Q$ is the midpoint of $AB$ $(\frac{a+b}{2}, 4)$,the slope of $AB$ is $\frac{5-3}{b-a} = \frac{2}{b-a}$. The slope of $OQ$ is $\frac{4-1}{\frac{a+b}{2}-1} = \frac{3}{\frac{a+b-2}{2}} = \frac{6}{a+b-2}$.
Since $OQ \perp AB$,$(\frac{2}{b-a}) \times (\frac{6}{a+b-2}) = -1 \Rightarrow 12 = -(b-a)(a+b-2) = (a-b)(a+b-2)$.
Substituting $b=-1$: $12 = (a+1)(a-3) = a^2 - 2a - 3$ $\Rightarrow a^2 - 2a - 15 = 0$ $\Rightarrow (a-5)(a+3) = 0$. So $a=5$ or $a=-3$.
Coordinates of $C(a, b)$ are $(5, -1)$ and $(-3, -1)$.
The distinct coordinates are $5, -3, -1$. The sum of their absolute values is $|5| + |-3| + |-1| = 5 + 3 + 1 = 9$.

(B) The lines are $L_1: x=2$,$L_2: y=3$,and $L_3: 4x+3y+7=0$.
The vertices are found by intersection:
$A = L_2 \cap L_3: y=3$ $\Rightarrow 4x+9+7=0$ $\Rightarrow 4x=-16$ $\Rightarrow x=-4$. So $A=(-4, 3)$.
$B = L_1 \cap L_2: x=2, y=3$. So $B=(2, 3)$.
$C = L_1 \cap L_3: x=2$ $\Rightarrow 8+3y+7=0$ $\Rightarrow 3y=-15$ $\Rightarrow y=-5$. So $C=(2, -5)$.
The side lengths are $c = AB = \sqrt{(2-(-4))^2 + (3-3)^2} = 6$,$a = BC = \sqrt{(2-2)^2 + (-5-3)^2} = 8$,and $b = AC = \sqrt{(2-(-4))^2 + (-5-3)^2} = \sqrt{6^2 + (-8)^2} = 10$.
The incentre $I = \left(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c}\right) = \left(\frac{8(-4)+10(2)+6(2)}{8+10+6}, \frac{8(3)+10(3)+6(-5)}{8+10+6}\right) = \left(\frac{-32+20+12}{24}, \frac{24+30-30}{24}\right) = (0, 1)$.
Since $\triangle ABC$ is a right-angled triangle at $B(2, 3)$,the circumcentre $S$ is the midpoint of the hypotenuse $AC$.
$S = \left(\frac{-4+2}{2}, \frac{3-5}{2}\right) = (-1, -1)$.
The distance $IS = \sqrt{(0-(-1))^2 + (1-(-1))^2} = \sqrt{1^2 + 2^2} = \sqrt{5}$.

(C) Let the vertices be $A(0, a)$,$B(2a, 0)$,and $C(x_1, y_1)$.
Since one of the sides is a horizontal line and it is not the $X$-axis,the side $AC$ must be horizontal.
Therefore,the $y$-coordinate of $C$ must be equal to the $y$-coordinate of $A$,so $y_1 = a$.
Since the triangle is isosceles and $AC$ is a side,we have $AC = BC$ or $AC = AB$ or $BC = AB$.
Given $AC$ is horizontal,$AC = \sqrt{(x_1 - 0)^2 + (a - a)^2} = |x_1|$.
$BC = \sqrt{(x_1 - 2a)^2 + (a - 0)^2} = \sqrt{(x_1 - 2a)^2 + a^2}$.
Setting $AC^2 = BC^2$,we get $x_1^2 = (x_1 - 2a)^2 + a^2$.
$x_1^2 = x_1^2 - 4ax_1 + 4a^2 + a^2$.
$4ax_1 = 5a^2$.
Since $a \neq 0$,$x_1 = \frac{5a}{4}$.
Thus,$x_1 + y_1 = \frac{5a}{4} + a = \frac{9a}{4}$.

(A) Line $L_1$ passes through $(2,1)$ and $(3, \frac{5}{2})$.
Slope of $L_1$ is $m_1 = \frac{\frac{5}{2} - 1}{3 - 2} = \frac{3/2}{1} = \frac{3}{2}$.
The equation of $L_1$ is $(y - 1) = \frac{3}{2}(x - 2)$,which simplifies to $3x - 2y = 4$.
Line $L_2$ is perpendicular to $L_1$,so its slope $m_2 = -\frac{1}{m_1} = -\frac{2}{3}$.
$L_2$ passes through $(4, -1)$,so its equation is $(y + 1) = -\frac{2}{3}(x - 4)$,which simplifies to $2x + 3y = 5$.
$L_1$ intersects the $y$-axis at $x=0$,so $-2y = 4 \Rightarrow y = -2$. Point is $(0, -2)$.
$L_2$ intersects the $y$-axis at $x=0$,so $3y = 5 \Rightarrow y = \frac{5}{3}$. Point is $(0, \frac{5}{3})$.
To find the intersection of $L_1$ and $L_2$,solve $3x - 2y = 4$ and $2x + 3y = 5$. Multiplying the first by $3$ and second by $2$: $9x - 6y = 12$ and $4x + 6y = 10$. Adding them gives $13x = 22 \Rightarrow x = \frac{22}{13}$. Substituting $x$ gives $y = \frac{7}{13}$.
The vertices of the triangle are $(0, -2)$,$(0, \frac{5}{3})$,and $(\frac{22}{13}, \frac{7}{13})$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times |\frac{5}{3} - (-2)| \times |\frac{22}{13}| = \frac{1}{2} \times \frac{11}{3} \times \frac{22}{13} = \frac{121}{39}$.

(C) Given the line equation: $\alpha x + \beta y + \sqrt{\alpha \beta} = 0$.
Dividing the equation by $-\sqrt{\alpha \beta}$,we get:
$\frac{\alpha x}{-\sqrt{\alpha \beta}} + \frac{\beta y}{-\sqrt{\alpha \beta}} = 1$
$\Rightarrow \frac{\sqrt{\alpha}}{\sqrt{\beta}} x + \frac{\sqrt{\beta}}{\sqrt{\alpha}} y = -1$.
The $x$-intercept is $a = -\sqrt{\frac{\beta}{\alpha}}$ and the $y$-intercept is $b = -\sqrt{\frac{\alpha}{\beta}}$.
The area of the triangle formed by the line with the coordinate axes is given by $Area = \frac{1}{2} |ab|$.
$Area = \frac{1}{2} |(-\sqrt{\frac{\beta}{\alpha}}) \times (-\sqrt{\frac{\alpha}{\beta}})| = \frac{1}{2} |1| = \frac{1}{2}$ square units.
Since the area is $\frac{1}{2}$,which is independent of $\alpha$ and $\beta$,the line forms a triangle of constant area with the coordinate axes.

(C) The slope of the perpendicular bisector of $AB$ is $m_1 = -\frac{2}{3}$. Thus,the slope of $AB$ is $m_{AB} = -\frac{1}{m_1} = \frac{3}{2}$.
Since $AB$ passes through $A(3,2)$,its equation is $y-2 = \frac{3}{2}(x-3) \Rightarrow 3x-2y-5=0$.
The intersection of $AB$ and its perpendicular bisector $2x+3y+1=0$ gives the midpoint $E$ of $AB$. Solving $3x-2y-5=0$ and $2x+3y+1=0$,we get $E(1,-1)$.
Let $B$ be $(x_1, y_1)$. Since $E$ is the midpoint of $AB$,$\frac{x_1+3}{2} = 1 \Rightarrow x_1 = -1$ and $\frac{y_1+2}{2} = -1 \Rightarrow y_1 = -4$. So,$B(-1,-4)$.
The slope of the perpendicular bisector of $AC$ is $m_2 = -\frac{1}{2}$. Thus,the slope of $AC$ is $m_{AC} = -\frac{1}{m_2} = 2$.
Since $AC$ passes through $A(3,2)$,its equation is $y-2 = 2(x-3) \Rightarrow 2x-y-4=0$.
The intersection of $AC$ and its perpendicular bisector $x+2y-12=0$ gives the midpoint $D$ of $AC$. Solving $2x-y-4=0$ and $x+2y-12=0$,we get $D(4,4)$.
Let $C$ be $(x_2, y_2)$. Since $D$ is the midpoint of $AC$,$\frac{x_2+3}{2} = 4 \Rightarrow x_2 = 5$ and $\frac{y_2+2}{2} = 4 \Rightarrow y_2 = 6$. So,$C(5,6)$.
The slope of $BC$ is $\frac{6-(-4)}{5-(-1)} = \frac{10}{6} = \frac{5}{3}$.

(A) The equation of the hypotenuse is $3x + 4y = 4$,which can be written as $y = -\frac{3}{4}x + 1$.
Thus,the slope of the hypotenuse is $m_1 = -\frac{3}{4}$.
Let the slopes of the other two sides be $m$.
Since it is an isosceles right-angled triangle,the angle between the hypotenuse and each of the other two sides is $45^{\circ}$.
Using the formula $\tan \theta = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$,we have:
$\tan 45^{\circ} = \left| \frac{m - (-3/4)}{1 + m(-3/4)} \right| = 1$.
$1 = \left| \frac{4m + 3}{4 - 3m} \right|$.
This gives two cases:
Case $1$: $\frac{4m + 3}{4 - 3m} = 1$ $\Rightarrow 4m + 3 = 4 - 3m$ $\Rightarrow 7m = 1$ $\Rightarrow m = \frac{1}{7}$.
Case $2$: $\frac{4m + 3}{4 - 3m} = -1$ $\Rightarrow 4m + 3 = -4 + 3m$ $\Rightarrow m = -7$.
Therefore,the slopes of the remaining two sides are $\frac{1}{7}$ and $-7$.

(A) To find the vertices of the triangle,we solve the equations of the lines in pairs:
$1$. For vertex $A$,solve $x+3y-4=0$ and $3x+y-4=0$:
Multiplying the first equation by $3$,we get $3x+9y-12=0$.
Subtracting $3x+y-4=0$ from this,we get $8y-8=0$,so $y=1$. Substituting $y=1$ into $x+3(1)-4=0$,we get $x=1$. Thus,$A = (1, 1)$.
$2$. For vertex $B$,solve $x+3y-4=0$ and $x+y-4=0$:
Subtracting the second from the first,we get $2y=0$,so $y=0$. Substituting $y=0$ into $x+y-4=0$,we get $x=4$. Thus,$B = (4, 0)$.
$3$. For vertex $C$,solve $3x+y-4=0$ and $x+y-4=0$:
Subtracting the second from the first,we get $2x=0$,so $x=0$. Substituting $x=0$ into $x+y-4=0$,we get $y=4$. Thus,$C = (0, 4)$.
Now,calculate the lengths of the sides:
$AB = \sqrt{(4-1)^2 + (0-1)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10}$
$BC = \sqrt{(0-4)^2 + (4-0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$
$CA = \sqrt{(1-0)^2 + (1-4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1+9} = \sqrt{10}$
Since $AB = CA = \sqrt{10}$,the triangle is an isosceles triangle.

(D) The equation of line $L$ passing through $(3,0)$ and $(1, 4/3)$ is given by:
$y - 0 = \frac{4/3 - 0}{1 - 3}(x - 3)$
$y = -\frac{2}{3}(x - 3)$ $\Rightarrow 3y = -2x + 6$ $\Rightarrow 2x + 3y = 6$ (Eq. $i$)
The slope of line $L$ is $m_1 = -2/3$. The slope of the line perpendicular to $L$ is $m_2 = -1/m_1 = 3/2$.
The equation of the line passing through $(8,1)$ with slope $3/2$ is:
$y - 1 = \frac{3}{2}(x - 8)$ $\Rightarrow 2y - 2 = 3x - 24$ $\Rightarrow 3x - 2y = 22$ (Eq. $ii$)
Solving Eq. $i$ and Eq. $ii$ for the intersection point:
$2x + 3y = 6$ (multiply by $2$) $\Rightarrow 4x + 6y = 12$
$3x - 2y = 22$ (multiply by $3$) $\Rightarrow 9x - 6y = 66$
Adding these gives $13x = 78 \Rightarrow x = 6$. Substituting $x=6$ into $2x + 3y = 6$ gives $12 + 3y = 6$ $\Rightarrow 3y = -6$ $\Rightarrow y = -2$. The intersection point is $(6, -2)$.
The line $2x + 3y = 6$ intersects the $Y$-axis $(x=0)$ at $(0, 2)$.
The line $3x - 2y = 22$ intersects the $Y$-axis $(x=0)$ at $(0, -11)$.
The vertices of the triangle are $(6, -2)$,$(0, 2)$,and $(0, -11)$.
The base of the triangle on the $Y$-axis is $|2 - (-11)| = 13$.
The height of the triangle is the absolute value of the $x$-coordinate of the intersection point,which is $|6| = 6$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 13 \times 6 = 39 \text{ sq units}$.

(B) Let the lines be $L_1: 7x+y-24=0$ and $L_2: x+7y-24=0$. The slopes are $m_1 = -7$ and $m_2 = -\frac{1}{7}$.
Since the lines are equal sides of an isosceles triangle,the third side makes equal angles with $L_1$ and $L_2$.
Let the slope of the third side be $m$. Then $\left| \frac{m - (-7)}{1 + m(-7)} \right| = \left| \frac{m - (-1/7)}{1 + m(-1/7)} \right|$.
$\left| \frac{m+7}{1-7m} \right| = \left| \frac{7m+1}{7-m} \right|$.
Case $1$: $\frac{m+7}{1-7m} = \frac{7m+1}{7-m}$ $\Rightarrow (m+7)(7-m) = (7m+1)(1-7m)$ $\Rightarrow 49-m^2 = 1-49m^2$ $\Rightarrow 48m^2 = -48$ (No real solution).
Case $2$: $\frac{m+7}{1-7m} = -\frac{7m+1}{7-m}$ $\Rightarrow (m+7)(7-m) = -(7m+1)(1-7m)$ $\Rightarrow 49-m^2 = -(1-49m^2-7m+7m) = 49m^2-1$ $\Rightarrow 50m^2 = 50$ $\Rightarrow m = \pm 1$.
For $m = -1$,the line passing through $(-1, 1)$ is $y-1 = -1(x+1)$ $\Rightarrow y-1 = -x-1$ $\Rightarrow x+y=0$.
For $m = 1$,the line passing through $(-1, 1)$ is $y-1 = 1(x+1)$ $\Rightarrow y-1 = x+1$ $\Rightarrow x-y+2=0$.

(B) Given,$ABCD$ is a rectangle.
Slope of $AB = \frac{-2-1}{3-2} = -3 = m_1$.
Since $AB \perp BC$,the slope of $BC$ is $m_2 = -\frac{1}{m_1} = \frac{1}{3}$.
Slope of $BC = \frac{b - (-2)}{a - 3} = \frac{b+2}{a-3} = \frac{1}{3}$.
$3(b+2) = a-3$ $\Rightarrow 3b + 6 = a - 3$ $\Rightarrow a - 3b = 9$ ... $(i)$.
Since $CD \parallel AB$,the slope of $CD$ is $m_3 = m_1 = -3$.
The points $C(a, b)$ and $P(3, 4)$ lie on the line $CD$.
Slope of $CP = \frac{4-b}{3-a} = -3$.
$4-b = -3(3-a)$ $\Rightarrow 4-b = -9 + 3a$ $\Rightarrow 3a + b = 13$ ... $(ii)$.
Multiplying $(ii)$ by $3$: $9a + 3b = 39$ ... $(iii)$.
Adding $(i)$ and $(iii)$: $(a - 3b) + (9a + 3b) = 9 + 39$ $\Rightarrow 10a = 48$ $\Rightarrow a = 4.8 = \frac{24}{5}$.
Substituting $a = 4.8$ in $(ii)$: $3(4.8) + b = 13$ $\Rightarrow 14.4 + b = 13$ $\Rightarrow b = -1.4 = -\frac{7}{5}$.
Now,$5a + 10b = 5(\frac{24}{5}) + 10(-\frac{7}{5}) = 24 - 14 = 10$.

(D) Let the rectangle be $ABCD$. The lines are $AB: 3x + y - 4 = 0$,$BC: x - ay - 10 = 0$,and $CD: bx + 2y + 9 = 0$.
Since $AB \parallel CD$,their slopes must be equal. Slope of $AB$ is $-3$. Slope of $CD$ is $-\frac{b}{2}$. Thus,$-\frac{b}{2} = -3 \Rightarrow b = 6$.
Since $AB \perp BC$,the product of their slopes is $-1$. Slope of $BC$ is $\frac{1}{a}$. Thus,$(-3) \times (\frac{1}{a}) = -1 \Rightarrow a = 3$.
Now,$AB: 3x + y - 4 = 0$ and $CD: 6x + 2y + 9 = 0$,which can be written as $3x + y + \frac{9}{2} = 0$.
The distance between parallel lines $AB$ and $CD$ is $BC = \frac{|-4 - 9/2|}{\sqrt{3^2 + 1^2}} = \frac{17/2}{\sqrt{10}} = \frac{17}{2\sqrt{10}}$.
The line $BC$ is $x - 3y - 10 = 0$. The side $CD$ is a line parallel to $AB$ and $AD$ is a line parallel to $BC$. The distance $CD$ is the distance between the parallel lines $AD$ and $BC$. Since $AD$ passes through $(1, 2)$,the distance $CD$ is the perpendicular distance from $(1, 2)$ to the line $BC: x - 3y - 10 = 0$.
$CD = \frac{|1 - 3(2) - 10|}{\sqrt{1^2 + (-3)^2}} = \frac{|1 - 6 - 10|}{\sqrt{10}} = \frac{15}{\sqrt{10}}$.
Area of rectangle $ABCD = BC \times CD = \frac{17}{2\sqrt{10}} \times \frac{15}{\sqrt{10}} = \frac{17 \times 15}{2 \times 10} = \frac{255}{20} = \frac{51}{4}$.

(B) Let the two given altitudes be $L_1: \sqrt{3}x - y + 8 - 4\sqrt{3} = 0$ and $L_2: \sqrt{3}x + y - 12 - 4\sqrt{3} = 0$.
The intersection of these two altitudes is the orthocenter of the triangle.
Adding $L_1$ and $L_2$: $(\sqrt{3}x - y + 8 - 4\sqrt{3}) + (\sqrt{3}x + y - 12 - 4\sqrt{3}) = 0$.
$2\sqrt{3}x - 4 - 8\sqrt{3} = 0$ $\Rightarrow 2\sqrt{3}x = 4 + 8\sqrt{3}$ $\Rightarrow x = \frac{2 + 4\sqrt{3}}{\sqrt{3}} = 4 + \frac{2}{\sqrt{3}}$.
Subtracting $L_1$ from $L_2$: $(\sqrt{3}x + y - 12 - 4\sqrt{3}) - (\sqrt{3}x - y + 8 - 4\sqrt{3}) = 0$.
$2y - 20 = 0 \Rightarrow y = 10$.
The orthocenter is $(4 + \frac{2}{\sqrt{3}}, 10)$.
In an equilateral triangle,the orthocenter is the same as the centroid.
Since the third altitude must pass through the orthocenter $(x_0, y_0) = (4 + \frac{2}{\sqrt{3}}, 10)$,we check the given options.
Substituting $(4 + \frac{2}{\sqrt{3}}, 10)$ into $y = 10$,we see it satisfies the equation.
Thus,the equation of the third altitude is $y = 10$.

(A) The equation of line $L_1$ passing through $A(3,4)$ with slope $m_1 = 1$ is:
$y - 4 = 1(x - 3) \implies x - y + 1 = 0$.
Let the slope of line $AC$ be $m$. The line $BC$ has the equation $2x - y + 4 = 0$,so its slope is $m_{BC} = 2$.
Since $AB = AC$,$\triangle ABC$ is an isosceles triangle with $A$ as the vertex. Thus,the base angles are equal: $\angle B = \angle C$.
The angle between $AC$ (slope $m$) and $BC$ (slope $2$) is equal to the angle between $AB$ (slope $1$) and $BC$ (slope $2$).
Using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$,we have:
$\left| \frac{m - 2}{1 + 2m} \right| = \left| \frac{1 - 2}{1 + (1)(2)} \right| = \left| \frac{-1}{3} \right| = \frac{1}{3}$.
This gives two cases:
$1) \frac{m - 2}{1 + 2m} = \frac{1}{3} \implies 3m - 6 = 1 + 2m \implies m = 7$.
$2) \frac{m - 2}{1 + 2m} = -\frac{1}{3} \implies 3m - 6 = -1 - 2m \implies 5m = 5 \implies m = 1$.
Since $m = 1$ corresponds to line $L_1$ (which is $AB$),the slope of $AC$ must be $m = 7$.
The equation of $AC$ passing through $A(3,4)$ with slope $7$ is:
$y - 4 = 7(x - 3) \implies y - 4 = 7x - 21 \implies 7x - y - 17 = 0$.

(D) Given lines are $L_1: \lambda x+4 y+2=0$,$L_2: 3 x+4 y-3=0$,$L_3: 2 x+\mu y+6=0$,$L_4: 2 x+y+3=0$.
Since $L_1 \parallel L_2$,their slopes must be equal: $-\frac{\lambda}{4} = -\frac{3}{4} \Rightarrow \lambda = 3$.
Since $L_3 \parallel L_4$,their slopes must be equal: $-\frac{2}{\mu} = -\frac{2}{1} \Rightarrow \mu = 1$.
The area of a parallelogram formed by lines $a_1 x+b_1 y+c_1=0$,$a_1 x+b_1 y+c_2=0$,$a_2 x+b_2 y+d_1=0$,and $a_2 x+b_2 y+d_2=0$ is given by $\frac{|(c_1-c_2)(d_1-d_2)|}{|a_1 b_2 - a_2 b_1|}$.
Here,$L_1: 3x+4y+2=0$,$L_2: 3x+4y-3=0$,$L_3: 2x+y+6=0$,$L_4: 2x+y+3=0$.
$c_1=2, c_2=-3, d_1=6, d_2=3$.
$a_1=3, b_1=4, a_2=2, b_2=1$.
Area $= \frac{|(2 - (-3))(6 - 3)|}{|(3)(1) - (2)(4)|} = \frac{|5 \times 3|}{|3 - 8|} = \frac{15}{|-5|} = \frac{15}{5} = 3$.
Thus,the area is $3$ square units.

(B) The given equation of the line is $3x - 4y = 6$.
Any line perpendicular to this line is of the form $4x + 3y = k$.
The intercepts of this line on the coordinate axes are $x = \frac{k}{4}$ and $y = \frac{k}{3}$.
The area of the triangle formed by the line with the coordinate axes is given by $\text{Area} = \frac{1}{2} \times |\text{base}| \times |\text{height}|$.
$\frac{1}{2} \times |\frac{k}{4}| \times |\frac{k}{3}| = 6$.
$|\frac{k^2}{24}| = 6$.
$k^2 = 144$.
$k = \pm 12$.
Thus,the required equations of the lines are $4x + 3y = 12$ or $4x + 3y = -12$.

(B) The given line is $2x - 3y + 7 = 0$.
Any line perpendicular to the given line is of the form $3x + 2y + k = 0$.
To find the intercepts,set $y = 0$: $3x + k = 0 \Rightarrow x = -\frac{k}{3}$.
Set $x = 0$: $2y + k = 0 \Rightarrow y = -\frac{k}{2}$.
The area of the triangle formed with the coordinate axes is given by $\frac{1}{2} |x_{intercept} \cdot y_{intercept}| = 3$.
$\frac{1}{2} |(-\frac{k}{3}) \cdot (-\frac{k}{2})| = 3$.
$\frac{1}{2} |\frac{k^2}{6}| = 3$.
$|k^2| = 36 \Rightarrow k = \pm 6$.
Substituting $k$ into the equation $3x + 2y + k = 0$,we get $3x + 2y = \pm 6$.