Points related to triangle Questions in English

(N/A) Given lines are:
$y=m_{1}x+c_{1}$ ..... $(1)$
$y=m_{2}x+c_{2}$ ..... $(2)$
$x=0$ ..... $(3)$
We know that the line $y=mx+c$ meets the line $x=0$ ($y$-axis) at the point $(0, c)$. Therefore,two vertices of the triangle formed by lines $(1)$,$(2)$,and $(3)$ are $P(0, c_{1})$ and $Q(0, c_{2})$.
The third vertex $R$ is obtained by solving equations $(1)$ and $(2)$:
$m_{1}x+c_{1} = m_{2}x+c_{2}$
$x(m_{1}-m_{2}) = c_{2}-c_{1}$
$x = \frac{c_{2}-c_{1}}{m_{1}-m_{2}}$
Substituting $x$ in $(1)$:
$y = m_{1}\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\right) + c_{1} = \frac{m_{1}c_{2}-m_{1}c_{1}+m_{1}c_{1}-m_{2}c_{1}}{m_{1}-m_{2}} = \frac{m_{1}c_{2}-m_{2}c_{1}}{m_{1}-m_{2}}$
So,$R = \left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}, \frac{m_{1}c_{2}-m_{2}c_{1}}{m_{1}-m_{2}}\right)$.
The area of the triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ is $\frac{1}{2}|x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})|$.
Here,vertices are $(0, c_{1}), (0, c_{2}), \left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}, \frac{m_{1}c_{2}-m_{2}c_{1}}{m_{1}-m_{2}}\right)$.
Area $= \frac{1}{2} |0(c_{2} - y_{R}) + 0(y_{R} - c_{1}) + \frac{c_{2}-c_{1}}{m_{1}-m_{2}}(c_{1}-c_{2})|$
Area $= \frac{1}{2} |\frac{-(c_{2}-c_{1})^{2}}{m_{1}-m_{2}}| = \frac{(c_{1}-c_{2})^{2}}{2|m_{1}-m_{2}|}$.

(C) Let the orthocentre be $H(x_0, y_0)$.
The slope of $BC$ is $m_{BC} = \frac{-5-1}{5-(-7)} = \frac{-6}{12} = -\frac{1}{2}$.
Since $AH \perp BC$,the slope of $AH$ is $m_{AH} = -\frac{1}{m_{BC}} = 2$.
The equation of altitude $AH$ passing through $A(-1, 7)$ is $y - 7 = 2(x + 1)$,which simplifies to $2x - y + 9 = 0$ ... $(1)$.
The slope of $AC$ is $m_{AC} = \frac{-5-7}{5-(-1)} = \frac{-12}{6} = -2$.
Since $BH \perp AC$,the slope of $BH$ is $m_{BH} = -\frac{1}{m_{AC}} = \frac{1}{2}$.
The equation of altitude $BH$ passing through $B(-7, 1)$ is $y - 1 = \frac{1}{2}(x + 7)$,which simplifies to $x - 2y + 9 = 0$ ... $(2)$.
Solving equations $(1)$ and $(2)$:
From $(1)$,$y = 2x + 9$. Substituting into $(2)$:
$x - 2(2x + 9) + 9 = 0$
$x - 4x - 18 + 9 = 0$
$-3x - 9 = 0 \Rightarrow x = -3$.
Then $y = 2(-3) + 9 = 3$.
Thus,the orthocentre is $(-3, 3)$.

(C) First,calculate the lengths of the sides of $\triangle ABC$:
$AB^2 = (1 - (-2))^2 + (9 - 3)^2 = 3^2 + 6^2 = 9 + 36 = 45 \Rightarrow AB = \sqrt{45}$
$BC^2 = (3 - 1)^2 + (8 - 9)^2 = 2^2 + (-1)^2 = 4 + 1 = 5 \Rightarrow BC = \sqrt{5}$
$AC^2 = (3 - (-2))^2 + (8 - 3)^2 = 5^2 + 5^2 = 25 + 25 = 50 \Rightarrow AC = \sqrt{50}$
Since $AB^2 + BC^2 = 45 + 5 = 50 = AC^2$,the triangle is a right-angled triangle with $\angle B = 90^{\circ}$.
In a right-angled triangle,the circum-center is the midpoint of the hypotenuse $AC$.
Circum-center $= \left(\frac{-2 + 3}{2}, \frac{3 + 8}{2}\right) = \left(\frac{1}{2}, \frac{11}{2}\right)$.
The midpoint of $BC$ is $\left(\frac{1 + 3}{2}, \frac{9 + 8}{2}\right) = \left(2, \frac{17}{2}\right)$.
The line $L$ passes through $\left(\frac{1}{2}, \frac{11}{2}\right)$ and $\left(2, \frac{17}{2}\right)$.
The slope $m = \frac{\frac{17}{2} - \frac{11}{2}}{2 - \frac{1}{2}} = \frac{3}{\frac{3}{2}} = 2$.
The equation of line $L$ is $y - \frac{11}{2} = 2(x - \frac{1}{2})$ $\Rightarrow y = 2x - 1 + \frac{11}{2}$ $\Rightarrow y = 2x + \frac{9}{2}$.
Since the line intersects the $y$-axis at $\left(0, \frac{\alpha}{2}\right)$,we have $\frac{\alpha}{2} = \frac{9}{2}$,which gives $\alpha = 9$.

(C) Let the sides be $AB: x - 2y + 1 = 0$ and $AC: 2x - y - 1 = 0$. Solving these,we get vertex $A(1, 1)$.
The altitude from vertex $B$ passes through $H\left(\frac{7}{3}, \frac{7}{3}\right)$ and is perpendicular to $AC$. The slope of $AC$ is $2$,so the slope of altitude $BH$ is $-\frac{1}{2}$. The equation of $BH$ is $y - \frac{7}{3} = -\frac{1}{2}\left(x - \frac{7}{3}\right)$,which simplifies to $x + 2y - 7 = 0$.
Vertex $B$ is the intersection of $AB$ and $BH$: $x - 2y = -1$ and $x + 2y = 7$. Adding gives $2x = 6 \Rightarrow x = 3$,so $y = 2$. Thus,$B(3, 2)$.
The altitude from vertex $C$ passes through $H\left(\frac{7}{3}, \frac{7}{3}\right)$ and is perpendicular to $AB$. The slope of $AB$ is $\frac{1}{2}$,so the slope of altitude $CH$ is $-2$. The equation of $CH$ is $y - \frac{7}{3} = -2\left(x - \frac{7}{3}\right)$,which simplifies to $2x + y - 7 = 0$.
Vertex $C$ is the intersection of $AC$ and $CH$: $2x - y = 1$ and $2x + y = 7$. Adding gives $4x = 8 \Rightarrow x = 2$,so $y = 3$. Thus,$C(2, 3)$.
The centroid $G$ of $\triangle ABC$ with vertices $A(1, 1), B(3, 2), C(2, 3)$ is $\left(\frac{1+3+2}{3}, \frac{1+2+3}{3}\right) = (2, 2)$.
The distance of the centroid $(2, 2)$ from the origin $(0, 0)$ is $\sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.

(C) In $\triangle PQR$,let the vertices be $P, Q, R$. Let $M$ be the midpoint of $QR$ and $N$ be the midpoint of $PR$. The medians $PM$ and $QN$ intersect at the centroid $G$.
Using the property of medians,$QG = \frac{2}{3}QN$ and $GM = \frac{1}{3}PM$.
In $\triangle QGM$,by the Pythagorean theorem,if $\angle QGM = 90^{\circ}$,then $QG^2 + GM^2 = QM^2$.
We know $QM = \frac{1}{2}QR$,so $QM^2 = \frac{1}{4}QR^2$.
Using Apollonius theorem for medians:
$QN^2 = \frac{2PQ^2 + 2QR^2 - PR^2}{4}$ and $PM^2 = \frac{2PQ^2 + 2PR^2 - QR^2}{4}$.
Then $QG^2 + GM^2 = \left(\frac{2}{3}QN\right)^2 + \left(\frac{1}{3}PM\right)^2 = \frac{4}{9}QN^2 + \frac{1}{9}PM^2$.
Substituting the expressions for $QN^2$ and $PM^2$:
$= \frac{4}{9} \left(\frac{2PQ^2 + 2QR^2 - PR^2}{4}\right) + \frac{1}{9} \left(\frac{2PQ^2 + 2PR^2 - QR^2}{4}\right)$
$= \frac{1}{9} \left( \frac{8PQ^2 + 8QR^2 - 4PR^2 + 2PQ^2 + 2PR^2 - QR^2}{4} \right)$
$= \frac{1}{9} \left( \frac{10PQ^2 + 7QR^2 - 2PR^2}{4} \right)$.
Given $PR^2 = 5PQ^2 - QR^2$,substitute this:
$= \frac{1}{9} \left( \frac{10PQ^2 + 7QR^2 - 2(5PQ^2 - QR^2)}{4} \right)$
$= \frac{1}{9} \left( \frac{10PQ^2 + 7QR^2 - 10PQ^2 + 2QR^2}{4} \right) = \frac{1}{9} \left( \frac{9QR^2}{4} \right) = \frac{1}{4}QR^2 = QM^2$.
Since $QG^2 + GM^2 = QM^2$,$\triangle QGM$ is a right-angled triangle with $\angle QGM = 90^{\circ}$.

(A) In $\triangle PQR$,let $PQ = r$,$QR = p$,and $PR = q$. Since $PQ < QR$,we have $r < p$.
$1$. The altitude $QQ_1$ is the shortest segment from $Q$ to the line $PR$. Thus,$PQ_1$ is the smallest distance.
$2$. The median $QQ_3$ bisects $PR$,so $PQ_3 = \frac{1}{2} PR$.
$3$. By the Angle Bisector Theorem,the angle bisector $QQ_2$ divides $PR$ in the ratio of the adjacent sides $PQ:QR = r:p$. Thus,$PQ_2 = \left(\frac{r}{r+p}\right) PR$.
Since $r < p$,we have $r+p > 2r$,which implies $\frac{r}{r+p} < \frac{r}{2r} = \frac{1}{2}$.
Therefore,$PQ_2 < \frac{1}{2} PR = PQ_3$.
Comparing the positions of the feet on $PR$,we have $PQ_1 < PQ_2 < PQ_3$. The correct option is $A$.

(D) Let the vertices be $A(1,2), B(2,3)$ and $C(3,1)$.
Let the orthocentre be $H(\alpha, \beta)$.
The slope of $AC$ is $m_{AC} = \frac{1-2}{3-1} = -\frac{1}{2}$.
Since $BH \perp AC$,the slope of $BH$ is $m_{BH} = 2$. Thus,$\frac{\beta-3}{\alpha-2} = 2$ $\Rightarrow \beta-3 = 2\alpha-4$ $\Rightarrow \beta = 2\alpha-1$.
The slope of $AB$ is $m_{AB} = \frac{3-2}{2-1} = 1$.
Since $CH \perp AB$,the slope of $CH$ is $m_{CH} = -1$. Thus,$\frac{\beta-1}{\alpha-3} = -1$ $\Rightarrow \beta-1 = -\alpha+3$ $\Rightarrow \beta = -\alpha+4$.
Equating the two expressions for $\beta$: $2\alpha-1 = -\alpha+4$ $\Rightarrow 3\alpha = 5$ $\Rightarrow \alpha = \frac{5}{3}$.
Then $\beta = 2(\frac{5}{3})-1 = \frac{7}{3}$.
The roots of the quadratic equation are $p = \alpha+4\beta = \frac{5}{3} + \frac{28}{3} = 11$ and $q = 4\alpha+\beta = \frac{20}{3} + \frac{7}{3} = 9$.
The quadratic equation is $(x-p)(x-q) = 0$ $\Rightarrow (x-11)(x-9) = 0$ $\Rightarrow x^2-20x+99 = 0$.

(B) First,find the vertices of the triangle by solving the equations in pairs:
$1$. $4x + 3y = 69$ and $4y - 3x = 17$: Multiplying the first by $4$ and second by $3$,we get $16x + 12y = 276$ and $-9x + 12y = 51$. Subtracting gives $25x = 225$,so $x = 9$. Then $36 + 3y = 69 \implies 3y = 33 \implies y = 11$. Vertex $A = (9, 11)$.
$2$. $4x + 3y = 69$ and $x + 7y = 61$: Multiplying the second by $4$,we get $4x + 28y = 244$. Subtracting the first gives $25y = 175$,so $y = 7$. Then $x + 49 = 61 \implies x = 12$. Vertex $B = (12, 7)$.
$3$. $4y - 3x = 17$ and $x + 7y = 61$: Multiplying the second by $3$,we get $3x + 21y = 183$. Adding to the first gives $25y = 200$,so $y = 8$. Then $x + 56 = 61 \implies x = 5$. Vertex $C = (5, 8)$.
Note that the lines $4x + 3y = 69$ and $4y - 3x = 17$ are perpendicular since $(4)(-3) + (3)(4) = 0$. Thus,the triangle is a right-angled triangle with the right angle at the intersection of these two lines,which is $(9, 11)$.
The circumcenter of a right-angled triangle is the midpoint of the hypotenuse. The hypotenuse is the line $x + 7y = 61$ connecting $(12, 7)$ and $(5, 8)$.
Circumcenter $(\alpha, \beta) = \left(\frac{12+5}{2}, \frac{7+8}{2}\right) = \left(\frac{17}{2}, \frac{15}{2}\right)$.
Now,calculate $(\alpha - \beta)^2 + \alpha + \beta = \left(\frac{17}{2} - \frac{15}{2}\right)^2 + \frac{17}{2} + \frac{15}{2} = (1)^2 + \frac{32}{2} = 1 + 16 = 17$.

(B) Let $A = (3, -7)$,$B = (-1, 2)$,and $C = (4, 5)$.
Slope of $BC = \frac{5 - 2}{4 - (-1)} = \frac{3}{5}$.
The altitude from $A$ to $BC$ is perpendicular to $BC$,so its slope is $-\frac{5}{3}$.
The equation of the altitude from $A$ is $y - (-7) = -\frac{5}{3}(x - 3)$,which simplifies to $3y + 21 = -5x + 15$,or $5x + 3y + 6 = 0$.
Slope of $AC = \frac{5 - (-7)}{4 - 3} = \frac{12}{1} = 12$.
The altitude from $B$ to $AC$ is perpendicular to $AC$,so its slope is $-\frac{1}{12}$.
The equation of the altitude from $B$ is $y - 2 = -\frac{1}{12}(x - (-1))$,which simplifies to $12y - 24 = -x - 1$,or $x + 12y = 23$.
Solving the system of equations:
$1) 5x + 3y = -6$
$2) x + 12y = 23 \implies x = 23 - 12y$
Substitute $x$ into equation $1$:
$5(23 - 12y) + 3y = -6$
$115 - 60y + 3y = -6$
$-57y = -121 \implies y = \frac{121}{57} = \beta$
$x = 23 - 12(\frac{121}{57}) = 23 - 4(\frac{121}{19}) = \frac{437 - 484}{19} = -\frac{47}{19} = \alpha$
Now,calculate $9\alpha - 6\beta + 60$:
$9(-\frac{47}{19}) - 6(\frac{121}{57}) + 60 = -\frac{423}{19} - \frac{2(121)}{19} + 60 = \frac{-423 - 242}{19} + 60 = -\frac{665}{19} + 60 = -35 + 60 = 25$.

(A) Let $P(6,1)$ be the orthocentre of $\triangle ABC$ with $A(5,-2)$,$B(8,3)$,and $C(h, k)$.
Since $AP \perp BC$,the slope of $AP = \frac{1 - (-2)}{6 - 5} = \frac{3}{1} = 3$.
Therefore,the slope of $BC = -\frac{1}{3}$.
The equation of line $BC$ passing through $B(8,3)$ is $y - 3 = -\frac{1}{3}(x - 8)$,which simplifies to $3y - 9 = -x + 8$,or $x + 3y - 17 = 0$.
Since $BP \perp AC$,the slope of $BP = \frac{1 - 3}{6 - 8} = \frac{-2}{-2} = 1$.
Therefore,the slope of $AC = -1$.
The equation of line $AC$ passing through $A(5,-2)$ is $y - (-2) = -1(x - 5)$,which simplifies to $y + 2 = -x + 5$,or $x + y - 3 = 0$.
To find $C(h, k)$,we solve the system:
$1) x + 3y = 17$
$2) x + y = 3$
Subtracting $(2)$ from $(1)$ gives $2y = 14$,so $y = 7$.
Substituting $y = 7$ into $(2)$ gives $x + 7 = 3$,so $x = -4$.
Thus,$C = (-4, 7)$.
Now,check which circle equation is satisfied by $C(-4, 7)$:
$(-4)^2 + (7)^2 = 16 + 49 = 65$.
So,$x^2 + y^2 - 65 = 0$ is the correct equation.

(B) The centroid $G$ of a triangle divides the line segment joining the orthocentre $H$ and circumcentre $O$ in the ratio $2:1$. Given $H = (-6, -8)$ and $O = (3, 4)$,the centroid $G$ is:
$G = \left( \frac{2(3) + 1(-6)}{2+1}, \frac{2(4) + 1(-8)}{2+1} \right) = (0, 0)$.
The lines $2x+3y-1=0$ and $x+2y-1=0$ intersect at $(-1, 1)$.
Since the orthocentre of the triangle is $(0, 0)$,the altitude from the vertex $(-1, 1)$ to the opposite side $ax+by-1=0$ passes through $(0, 0)$.
The slope of the line joining $(-1, 1)$ and $(0, 0)$ is $m_1 = \frac{0-1}{0-(-1)} = -1$.
The slope of the line $ax+by-1=0$ is $m_2 = -a/b$.
Since the altitude is perpendicular to the side,$m_1 \times m_2 = -1$ $\Rightarrow (-1) \times (-a/b) = -1$ $\Rightarrow a/b = -1$ $\Rightarrow a = -b$.
The side is $ax - ay - 1 = 0$.
Since the orthocentre $(0, 0)$ is the intersection of altitudes,the altitude from the intersection of $2x+3y-1=0$ and $ax-ay-1=0$ must pass through $(0, 0)$.
Solving $2x+3y=1$ and $ax-ay=1$ gives the vertex $V = \left( \frac{a+3}{5a}, \frac{a-2}{5a} \right)$.
The altitude from $V$ to $x+2y-1=0$ (slope $-1/2$) has slope $2$. The line passing through $(0, 0)$ with slope $2$ is $y=2x$.
Substituting $V$ into $y=2x$: $\frac{a-2}{5a} = 2 \left( \frac{a+3}{5a} \right)$ $\Rightarrow a-2 = 2a+6$ $\Rightarrow a = -8$.
Thus $b = 8$,and $|a-b| = |-8-8| = 16$.

(C) The slope of $AB$ is $M_{AB} = \frac{3 - (-1)}{-2 - 3} = \frac{4}{-5}$.
Since $CP \perp AB$,the slope of $CP$ is $M_{CP} = \frac{5}{4}$.
The equation of line $CP$ passing through $P(1, 1)$ is $y - 1 = \frac{5}{4}(x - 1) \Rightarrow 5x - 4y - 1 = 0$ ... $(1)$.
The slope of $AP$ is $M_{AP} = \frac{1 - (-1)}{1 - 3} = \frac{2}{-2} = -1$.
Since $BC \perp AP$,the slope of $BC$ is $M_{BC} = 1$.
The equation of line $BC$ passing through $B(-2, 3)$ is $y - 3 = 1(x + 2) \Rightarrow x - y + 5 = 0$ ... $(2)$.
Solving $(1)$ and $(2)$: From $(2)$,$y = x + 5$. Substituting into $(1)$: $5x - 4(x + 5) - 1 = 0$ $\Rightarrow 5x - 4x - 20 - 1 = 0$ $\Rightarrow x = 21$. Thus $\alpha = 21$.
Then $\beta = 21 + 5 = 26$. So,$\alpha + \beta = 21 + 26 = 47$.
For the circumcentre $(h, k)$ of $\triangle PAB$,note that $P$ is the orthocentre of $\triangle ABC$. $A$ known property is that the circumcircle of $\triangle PAB$ has the same radius as the circumcircle of $\triangle ABC$,and its centre is the reflection of the circumcentre of $\triangle ABC$ across $AB$. Alternatively,the circumcentre of $\triangle PAB$ is the point $O'$ such that $O'A = O'B = O'P$. The perpendicular bisector of $AB$ is $y - 1 = \frac{5}{4}(x - 0.5)$. The perpendicular bisector of $AP$ is $y - 0 = 1(x - 2)$. Solving these gives $h = -19/2$ and $k = -23/2$.
Thus,$2(h + k) = 2(-19/2 - 23/2) = -42$.
The final value is $(\alpha + \beta) + 2(h + k) = 47 - 42 = 5$.

(C) point $R$ inside a triangle $OPQ$ such that the areas of triangles $OPR, PQR, OQR$ are equal is the centroid of the triangle.
Given the vertices $O(0,0), P(3,4), Q(6,0)$,the coordinates of the centroid $R(x, y)$ are given by:
$x = \frac{x_1 + x_2 + x_3}{3} = \frac{0 + 3 + 6}{3} = \frac{9}{3} = 3$
$y = \frac{y_1 + y_2 + y_3}{3} = \frac{0 + 4 + 0}{3} = \frac{4}{3}$
Thus,the coordinates of $R$ are $\left(3, \frac{4}{3}\right)$.

(B) The vertices $A(6, 8)$,$B(10 \cos \alpha, -10 \sin \alpha)$,and $C(-10 \sin \alpha, 10 \cos \alpha)$ all lie on the circle $x^2 + y^2 = 100$. Thus,the circumcenter $O$ is $(0, 0)$.
In any triangle,the orthocenter $L$,centroid $G$,and circumcenter $O$ are collinear,and $G$ divides $OL$ in the ratio $2:1$.
Using the section formula,$G(h, k) = \left(\frac{1 \cdot a + 2 \cdot 0}{1+2}, \frac{1 \cdot 9 + 2 \cdot 0}{1+2}\right) = \left(\frac{a}{3}, 3\right)$.
Thus,$h = \frac{a}{3} \Rightarrow a = 3h$ and $k = 3$.
Also,the centroid $G(h, k)$ is given by $\left(\frac{6 + 10 \cos \alpha - 10 \sin \alpha}{3}, \frac{8 - 10 \sin \alpha + 10 \cos \alpha}{3}\right) = (h, k)$.
Since $k = 3$,we have $\frac{8 + 10(\cos \alpha - \sin \alpha)}{3} = 3 \Rightarrow 10(\cos \alpha - \sin \alpha) = 1$.
Squaring both sides,$100(\cos^2 \alpha + \sin^2 \alpha - 2 \sin \alpha \cos \alpha) = 1$ $\Rightarrow 100(1 - \sin 2\alpha) = 1$ $\Rightarrow 100 \sin 2\alpha = 99$.
From $h = \frac{6 + 10(\cos \alpha - \sin \alpha)}{3}$,we get $h = \frac{6 + 1}{3} = \frac{7}{3}$.
Now,calculate $5a - 3h + 6k + 100 \sin 2\alpha = 5(3h) - 3h + 6(3) + 99 = 12h + 18 + 99 = 12(\frac{7}{3}) + 117 = 28 + 117 = 145$.

(B) The altitude from $P$ to $QR$ is a vertical line since $P$ and $Q$ have the same $y$-coordinate $(y=4)$. The line $PQ$ is horizontal. The altitude from $P$ to $QR$ passes through $P(5, 4)$ and the orthocenter $O\left(2, \frac{14}{5}\right)$. Since $QR$ is perpendicular to $PO$,the slope of $PO$ is $m_{PO} = \frac{4 - 14/5}{5 - 2} = \frac{6/5}{3} = \frac{2}{5}$. Thus,the slope of $QR$ is $m_{QR} = -\frac{5}{2}$.
Since $Q(-2, 4)$ lies on $QR$,the equation of $QR$ is $y - 4 = -\frac{5}{2}(x + 2) \implies 2y - 8 = -5x - 10 \implies 5x + 2y + 2 = 0$.
The altitude from $Q$ to $PR$ passes through $Q(-2, 4)$ and $O\left(2, \frac{14}{5}\right)$. The slope of $QO$ is $m_{QO} = \frac{14/5 - 4}{2 - (-2)} = \frac{-6/5}{4} = -\frac{3}{10}$. Thus,the slope of $PR$ is $m_{PR} = \frac{10}{3}$.
The equation of $PR$ is $y - 4 = \frac{10}{3}(x - 5) \implies 3y - 12 = 10x - 50 \implies 10x - 3y - 38 = 0$.
Solving $5x + 2y = -2$ and $10x - 3y = 38$: Multiply the first by $2$: $10x + 4y = -4$. Subtracting: $7y = -42 \implies y = -6$. Then $5x + 2(-6) = -2 \implies 5x = 10 \implies x = 2$. So $R(2, -6)$.
The area is $\frac{1}{2} |x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)| = 35 \implies \frac{1}{2} |5(4 - (-6)) + (-2)(-6 - 4) + 2(4 - 4)| = 35 \implies \frac{1}{2} |50 + 20 + 0| = 35$,which is consistent.
The centroid $C(c, d) = \left(\frac{5 - 2 + 2}{3}, \frac{4 + 4 - 6}{3}\right) = \left(\frac{5}{3}, \frac{2}{3}\right)$.
Therefore,$c + 2d = \frac{5}{3} + 2\left(\frac{2}{3}\right) = \frac{5+4}{3} = 3$.

(B) The median through vertex $A$ passes through the midpoint $M$ of the side $BC$.
Midpoint $M$ of $BC$ is given by $(\frac{2-4}{2}, \frac{1+5}{2}) = (-1, 3)$.
The median passes through $A(3, k)$ and $M(-1, 3)$.
The slope of the median $AM$ is $m = \frac{3-k}{-1-3} = \frac{3-k}{-4}$.
The equation of the line $AM$ is $y - 3 = \frac{3-k}{-4}(x + 1)$.
$-4y + 12 = (3-k)x + (3-k)$.
$(3-k)x + 4y = 9+k$.
Comparing this with the given equation $x + 4y = p$,we equate the coefficients of $x$ and $y$.
Since the coefficient of $y$ is $4$ in both,we have $3-k = 1$,which gives $k = 2$.
Thus,$k = 2$.

(B) The vertices are $A(0,0)$,$B(3,0)$,and $C(0,-4)$.
Calculate the side lengths:
$c = AB = \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{9} = 3$
$b = AC = \sqrt{(0-0)^2 + (-4-0)^2} = \sqrt{16} = 4$
$a = BC = \sqrt{(0-3)^2 + (-4-0)^2} = \sqrt{9+16} = \sqrt{25} = 5$
The coordinates of the incentre $(I)$ are given by the formula:
$I = \left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c} \right)$
Here,$(x_1, y_1) = (0,0)$,$(x_2, y_2) = (3,0)$,and $(x_3, y_3) = (0,-4)$.
$I = \left( \frac{5(0) + 4(3) + 3(0)}{5+4+3}, \frac{5(0) + 4(0) + 3(-4)}{5+4+3} \right)$
$I = \left( \frac{12}{12}, \frac{-12}{12} \right) = (1, -1)$.

(D) Let the vertices of the triangle be $A(-2,3)$,$B(6,-1)$,and $C(4,3)$. Let $F(x,y)$ be the circumcentre. The circumcentre is equidistant from the vertices,so $FA^2 = FB^2 = FC^2$.
$FA^2 = (x+2)^2 + (y-3)^2$
$FB^2 = (x-6)^2 + (y+1)^2$
$FC^2 = (x-4)^2 + (y-3)^2$
Equating $FA^2 = FC^2$:
$(x+2)^2 + (y-3)^2 = (x-4)^2 + (y-3)^2$
$(x+2)^2 = (x-4)^2$
$x^2 + 4x + 4 = x^2 - 8x + 16$
$12x = 12 \implies x = 1$
Equating $FA^2 = FB^2$:
$(1+2)^2 + (y-3)^2 = (1-6)^2 + (y+1)^2$
$9 + y^2 - 6y + 9 = 25 + y^2 + 2y + 1$
$18 - 6y = 26 + 2y$
$-8 = 8y \implies y = -1$
Thus,the coordinates of the circumcentre are $(1,-1)$.

(C) The circumcentre $C$ divides the line segment joining the orthocentre $A(-3, 5)$ and the centroid $B(3, 3)$ in the ratio $2:1$ externally,such that $B$ lies between $A$ and $C$.
Using the section formula for external division,the coordinates of $C(x, y)$ are given by:
$x = \frac{2(3) - 1(-3)}{2 - 1} = \frac{6 + 3}{1} = 9$
$y = \frac{2(3) - 1(5)}{2 - 1} = \frac{6 - 5}{1} = 1$
So,$C = (9, 1)$.
The diameter of the circle is the length of the segment $AC$.
$AC = \sqrt{(9 - (-3))^2 + (1 - 5)^2} = \sqrt{12^2 + (-4)^2} = \sqrt{144 + 16} = \sqrt{160} = 4\sqrt{10}$.
The radius of the circle is $\frac{1}{2} AC = \frac{1}{2} \times 4\sqrt{10} = 2\sqrt{10}$.

(B) The median through vertex $R$ bisects the side $PQ$ at point $M$.
$M$ is the midpoint of $PQ$,so $M = \left( \frac{2 + (-2)}{2}, \frac{2 + 4}{2} \right) = (0, 3)$.
The median passes through $R(3, 4)$ and $M(0, 3)$.
The slope of the line $RM$ is $m = \frac{3 - 4}{0 - 3} = \frac{-1}{-3} = \frac{1}{3}$.
The equation of the line is $y - y_1 = m(x - x_1)$.
$y - 3 = \frac{1}{3}(x - 0)$
$3(y - 3) = x$
$3y - 9 = x$
$x - 3y + 9 = 0$.

(D) 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 $F(11, 9)$,$E(2, 1)$,and $D(2, -1)$.
Since the centroid of a triangle formed by joining the mid-points of the sides of a triangle is the same as the centroid of the original triangle,we can directly calculate it.
The centroid $(G)$ of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
Let the mid-points be $(x_a, y_a) = (11, 9)$,$(x_b, y_b) = (2, 1)$,and $(x_c, y_c) = (2, -1)$.
The centroid of the triangle formed by these mid-points is $\left(\frac{11+2+2}{3}, \frac{9+1-1}{3}\right) = \left(\frac{15}{3}, \frac{9}{3}\right) = (5, 3)$.
Thus,the centroid of the original triangle is $(5, 3)$.

(B) Let the vertices of the triangle be $A(6,0), B(0,6)$ and $C(6,6)$.
First,we calculate the lengths of the sides:
$AB = \sqrt{(0-6)^2 + (6-0)^2} = \sqrt{36+36} = 6\sqrt{2}$
$BC = \sqrt{(6-0)^2 + (6-6)^2} = \sqrt{36+0} = 6$
$CA = \sqrt{(6-6)^2 + (0-6)^2} = \sqrt{0+36} = 6$
Since $AB^2 = BC^2 + CA^2$ $(72 = 36 + 36)$,the triangle is a right-angled triangle at vertex $C(6,6)$.
For a right-angled triangle,the circumcentre is the midpoint of the hypotenuse $AB$.
Circumcentre $O = \left(\frac{6+0}{2}, \frac{0+6}{2}\right) = (3,3)$.
The centroid $G$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{6+0+6}{3}, \frac{0+6+6}{3}\right) = (4,4)$.
The distance between the circumcentre $(3,3)$ and the centroid $(4,4)$ is $\sqrt{(4-3)^2 + (4-3)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.

(A) The vertices of the triangle are $A \equiv (2,3)$,$B \equiv (8,10)$,and $C \equiv (5,5)$.
The formula for the centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
Substituting the given values:
$G = \left(\frac{2+8+5}{3}, \frac{3+10+5}{3}\right)$
$G = \left(\frac{15}{3}, \frac{18}{3}\right)$
$G = (5, 6)$

(B) The centroid $G$ of a triangle $ABC$ divides the median $AD$ in the ratio $2:1$.
Let $A = (2, 3)$,$G = (7, 5)$,and $D = (x, y)$.
Using the section formula,the coordinates of $G$ are given by:
$G = \left(\frac{2 \cdot x + 1 \cdot 2}{2+1}, \frac{2 \cdot y + 1 \cdot 3}{2+1}\right)$
Equating the coordinates:
$7 = \frac{2x + 2}{3}$ $\Rightarrow 21 = 2x + 2$ $\Rightarrow 2x = 19$ $\Rightarrow x = \frac{19}{2}$
$5 = \frac{2y + 3}{3}$ $\Rightarrow 15 = 2y + 3$ $\Rightarrow 2y = 12$ $\Rightarrow y = 6$
Therefore,the coordinates of point $D$ are $\left(\frac{19}{2}, 6\right)$.

(D) The given vertices are $A(0,0)$,$B(0, \frac{3}{2})$,and $C(-5,0)$.
Since $A$ is at the origin $(0,0)$,$AB$ lies on the $y$-axis and $AC$ lies on the $x$-axis.
This means the triangle is a right-angled triangle with the right angle at vertex $A(0,0)$.
In a right-angled triangle,the orthocentre is the vertex where the right angle is located.
Therefore,the orthocentre of $\triangle ABC$ is $(0,0)$.

(B) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.
Since $AD$ is the median,$D$ is the midpoint of $BC$,so $D = (\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2})$.
$E$ is the midpoint of $AD$,so $E = (\frac{x_1 + \frac{x_2+x_3}{2}}{2}, \frac{y_1 + \frac{y_2+y_3}{2}}{2}) = (\frac{2x_1+x_2+x_3}{4}, \frac{2y_1+y_2+y_3}{4})$.
Using the property of the centroid $G$ of triangle $ABC$,$G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.
By applying Menelaus' theorem to triangle $ADC$ with transversal $B-E-F$,we find that $F$ divides $AC$ in the ratio $1: 2$.
Thus,$AF: FC = 1: 2$,which implies $AF: AC = 1: (1+2) = 1: 3$.

(A) Let the vertices be $A(2,2)$,$B(5,1)$,and $C(4,4)$.
To find the orthocentre,we find the equations of two altitudes.
Slope of $BC = \frac{4-1}{4-5} = \frac{3}{-1} = -3$.
The altitude from $A(2,2)$ to $BC$ has slope $m_1 = -\frac{1}{-3} = \frac{1}{3}$.
Equation of altitude from $A$: $y-2 = \frac{1}{3}(x-2)$ $\Rightarrow 3y-6 = x-2$ $\Rightarrow x-3y = -4 \quad \dots(i)$.
Slope of $AC = \frac{4-2}{4-2} = \frac{2}{2} = 1$.
The altitude from $B(5,1)$ to $AC$ has slope $m_2 = -\frac{1}{1} = -1$.
Equation of altitude from $B$: $y-1 = -1(x-5)$ $\Rightarrow y-1 = -x+5$ $\Rightarrow x+y = 6 \quad \dots(ii)$.
Adding $(i)$ and $(ii)$ multiplied by $3$: $(x-3y) + 3(x+y) = -4 + 18$ $\Rightarrow 4x = 14$ $\Rightarrow x = \frac{7}{2}$.
Substituting $x = \frac{7}{2}$ in $(ii)$: $\frac{7}{2} + y = 6 \Rightarrow y = 6 - \frac{7}{2} = \frac{5}{2}$.
Thus,the orthocentre $(\alpha, \beta) = (\frac{7}{2}, \frac{5}{2})$.
Therefore,$\alpha+\beta = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6$.

(A) Let the vertices of the triangle be $A(-2, 3)$,$B(2, -1)$,and $C(4, 0)$.
Let $O(x, y)$ be the circumcentre of the triangle.
By definition,the circumcentre is equidistant from all vertices,so $OA = OB = OC$,which implies $OA^2 = OB^2 = OC^2$.
First,set $OA^2 = OB^2$:
$(x + 2)^2 + (y - 3)^2 = (x - 2)^2 + (y + 1)^2$
$x^2 + 4x + 4 + y^2 - 6y + 9 = x^2 - 4x + 4 + y^2 + 2y + 1$
$8x - 8y = -8 \Rightarrow x - y = -1$ ... $(i)$
Next,set $OB^2 = OC^2$:
$(x - 2)^2 + (y + 1)^2 = (x - 4)^2 + (y - 0)^2$
$x^2 - 4x + 4 + y^2 + 2y + 1 = x^2 - 8x + 16 + y^2$
$4x + 2y = 11$ ... $(ii)$
Solving equations $(i)$ and $(ii)$:
From $(i)$,$y = x + 1$.
Substitute into $(ii)$: $4x + 2(x + 1) = 11$ $\Rightarrow 6x = 9$ $\Rightarrow x = \frac{3}{2}$.
Then $y = \frac{3}{2} + 1 = \frac{5}{2}$.
Thus,the circumcentre is $\left(\frac{3}{2}, \frac{5}{2}\right)$.

(B) Let the equations of the medians through $B$ and $C$ be $L_1: x+y=5$ and $L_2: x=4$ respectively. The point of intersection of these medians is the centroid $G$. Solving $x+y=5$ and $x=4$,we get $G=(4, 1)$.
Let $C$ lie on $x=4$,so $C=(4, y_C)$. Let $B$ lie on $x+y=5$,so $B=(x_B, 5-x_B)$.
Using the centroid formula $G = (\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3})$:
$4 = \frac{1+x_B+4}{3}$ $\Rightarrow 12 = 5+x_B$ $\Rightarrow x_B=7$. Thus $B=(7, 5-7) = (7, -2)$.
$1 = \frac{2+y_B+y_C}{3}$ $\Rightarrow 3 = 2+(-2)+y_C$ $\Rightarrow y_C=3$. Thus $C=(4, 3)$.
The vertices are $A(1, 2)$,$B(7, -2)$,and $C(4, 3)$.
The area of $\triangle ABC = \frac{1}{2} |x_A(y_B-y_C) + x_B(y_C-y_A) + x_C(y_A-y_B)|$
$= \frac{1}{2} |1(-2-3) + 7(3-2) + 4(2-(-2))|$
$= \frac{1}{2} |-5 + 7 + 16| = \frac{1}{2} |18| = 9$.

(A) Let the vertices of the triangle be $O(0,0)$,$B(4,0)$,and $C(3,4)$.
To find the orthocenter,we need the intersection of the altitudes.
$1$. The altitude from $C(3,4)$ to the side $OB$ (which lies on the $x$-axis) is a vertical line passing through $x=3$. Thus,the equation of this altitude is $x=3$.
$2$. The slope of side $BC$ is $m_{BC} = \frac{4-0}{3-4} = \frac{4}{-1} = -4$.
$3$. The altitude from $O(0,0)$ to $BC$ is perpendicular to $BC$. Its slope is $m_{\perp} = -\frac{1}{m_{BC}} = -\frac{1}{-4} = \frac{1}{4}$.
$4$. The equation of the altitude from $O$ is $y - 0 = \frac{1}{4}(x - 0)$,which simplifies to $y = \frac{1}{4}x$.
$5$. Substituting $x=3$ into the equation $y = \frac{1}{4}x$,we get $y = \frac{1}{4}(3) = \frac{3}{4}$.
Therefore,the orthocenter is $\left(3, \frac{3}{4}\right)$.

(D) Let the vertices be $A(1, \sqrt{3})$,$B(-1, -\sqrt{3})$,and $C(3, -\sqrt{3})$.
Calculate the side lengths:
$AB = \sqrt{(-1-1)^2 + (-\sqrt{3}-\sqrt{3})^2} = \sqrt{(-2)^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = 4$.
$BC = \sqrt{(3 - (-1))^2 + (-\sqrt{3} - (-\sqrt{3}))^2} = \sqrt{4^2 + 0^2} = 4$.
$AC = \sqrt{(3-1)^2 + (-\sqrt{3}-\sqrt{3})^2} = \sqrt{2^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = 4$.
Since $AB = BC = AC = 4$,the triangle is an equilateral triangle.
In an equilateral triangle,the circumcentre coincides with the centroid.
The centroid $(G)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
$G = \left(\frac{1-1+3}{3}, \frac{\sqrt{3}-\sqrt{3}-\sqrt{3}}{3}\right) = \left(\frac{3}{3}, \frac{-\sqrt{3}}{3}\right) = \left(1, -\frac{1}{\sqrt{3}}\right)$.

(C) Let the vertices be $A(-2,-1)$,$B(6,-1)$,and $C(2,5)$.
First,find the perpendicular bisector of $AB$. Since $A$ and $B$ have the same $y$-coordinate,the line $AB$ is horizontal. The midpoint is $(\frac{-2+6}{2}, -1) = (2,-1)$. The perpendicular bisector is the vertical line $x=2$ ... $(i)$.
Next,find the perpendicular bisector of $BC$. The midpoint of $BC$ is $(\frac{6+2}{2}, \frac{-1+5}{2}) = (4,2)$. The slope of $BC$ is $m = \frac{5-(-1)}{2-6} = \frac{6}{-4} = -\frac{3}{2}$. The slope of the perpendicular bisector is $\frac{2}{3}$. The equation is $y-2 = \frac{2}{3}(x-4)$ $\Rightarrow 3y-6 = 2x-8$ $\Rightarrow 2x-3y=2$ ... (ii).
Substitute $x=2$ from $(i)$ into (ii): $2(2)-3y=2$ $\Rightarrow 4-3y=2$ $\Rightarrow 3y=2$ $\Rightarrow y=\frac{2}{3}$.
The circumcentre is $(2, \frac{2}{3})$.
The quadratic equation with roots $2$ and $\frac{2}{3}$ is $x^2 - (2+\frac{2}{3})x + 2(\frac{2}{3}) = 0$.
$x^2 - \frac{8}{3}x + \frac{4}{3} = 0 \Rightarrow 3x^2-8x+4=0$.
Thus,option $(c)$ is correct.

(C) Let the vertices of the triangle be $A(h, k)$,$B(5, -1)$,and $C(-2, 3)$. The origin $O(0, 0)$ is the orthocentre.
Since $AO \perp BC$,the slope of $AO \times$ slope of $BC = -1$.
Slope of $BC = \frac{3 - (-1)}{-2 - 5} = \frac{4}{-7}$.
Slope of $AO = \frac{k - 0}{h - 0} = \frac{k}{h}$.
Thus,$\frac{k}{h} \times \left(-\frac{4}{7}\right) = -1$ $\Rightarrow \frac{k}{h} = \frac{7}{4}$ $\Rightarrow 7h - 4k = 0$ (Eq. $1$).
Since $BO \perp AC$,the slope of $BO \times$ slope of $AC = -1$.
Slope of $BO = \frac{-1 - 0}{5 - 0} = -\frac{1}{5}$.
Slope of $AC = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$.
Thus,$\left(-\frac{1}{5}\right) \times \left(\frac{k - 3}{h + 2}\right) = -1$ $\Rightarrow k - 3 = 5(h + 2)$ $\Rightarrow 5h - k + 13 = 0$ (Eq. $2$).
From Eq. $1$,$k = \frac{7h}{4}$. Substituting this into Eq. $2$:
$5h - \frac{7h}{4} + 13 = 0$ $\Rightarrow \frac{20h - 7h}{4} = -13$ $\Rightarrow 13h = -52$ $\Rightarrow h = -4$.
Then $k = \frac{7(-4)}{4} = -7$.
Therefore,the third vertex is $(-4, -7)$.

(B) Let the vertices be $A(-2, 3), B(2, -1)$,and $C(4, 0)$.
First,find the orthocentre $H$. The slope of $BC = \frac{0 - (-1)}{4 - 2} = \frac{1}{2}$. The altitude from $A$ is perpendicular to $BC$,so its slope is $-2$. The equation is $y - 3 = -2(x + 2) \Rightarrow 2x + y + 1 = 0$.
The slope of $AC = \frac{0 - 3}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2}$. The altitude from $B$ is perpendicular to $AC$,so its slope is $2$. The equation is $y - (-1) = 2(x - 2)$ $\Rightarrow y + 1 = 2x - 4$ $\Rightarrow 2x - y - 5 = 0$.
Solving $2x + y + 1 = 0$ and $2x - y - 5 = 0$ by adding them: $4x - 4 = 0 \Rightarrow x = 1$. Substituting $x=1$ into $2x + y + 1 = 0$ gives $2(1) + y + 1 = 0 \Rightarrow y = -3$. Thus,the orthocentre $H = (1, -3)$.
Next,find the centroid $G = \left(\frac{-2+2+4}{3}, \frac{3-1+0}{3}\right) = \left(\frac{4}{3}, \frac{2}{3}\right)$.
The equation of the line joining $H(1, -3)$ and $G\left(\frac{4}{3}, \frac{2}{3}\right)$ is given by $y - y_1 = m(x - x_1)$,where $m = \frac{\frac{2}{3} - (-3)}{\frac{4}{3} - 1} = \frac{\frac{11}{3}}{\frac{1}{3}} = 11$.
So,$y - (-3) = 11(x - 1)$ $\Rightarrow y + 3 = 11x - 11$ $\Rightarrow 11x - y - 14 = 0$.

(B) The vertices of the triangle are $A(2,3)$,$B(2,4)$,and $C(4,3)$.
Since $AB$ is vertical (along $x=2$) and $AC$ is horizontal (along $y=3$),the triangle is a right-angled triangle with the right angle at $A$.
For a right-angled triangle,the orthocentre $O$ is the vertex at the right angle. Thus,$O = A = (2,3)$.
Therefore,$AO^2 = (2-2)^2 + (3-3)^2 = 0$.
The centroid $G$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{2+2+4}{3}, \frac{3+4+3}{3}\right) = \left(\frac{8}{3}, \frac{10}{3}\right)$.
We calculate $9BG^2 = 9 \times \left[ (2 - \frac{8}{3})^2 + (4 - \frac{10}{3})^2 \right] = 9 \times [(-\frac{2}{3})^2 + (\frac{2}{3})^2] = 9 \times [\frac{4}{9} + \frac{4}{9}] = 9 \times \frac{8}{9} = 8$.
The centroid $G$ divides the line segment joining the orthocentre $O$ and the circumcentre $S$ in the ratio $2:1$. Using the section formula,$G = \frac{1 \cdot O + 2 \cdot S}{3}$ $\Rightarrow 3G = O + 2S$ $\Rightarrow S = \frac{3G - O}{2}$.
$S = \frac{3(\frac{8}{3}, \frac{10}{3}) - (2,3)}{2} = \frac{(8-2, 10-3)}{2} = \frac{(6, 7)}{2} = (3, \frac{7}{2})$.
Now,$4CS^2 = 4 \times [(4-3)^2 + (3-\frac{7}{2})^2] = 4 \times [1^2 + (-\frac{1}{2})^2] = 4 \times [1 + \frac{1}{4}] = 4 \times \frac{5}{4} = 5$.
Finally,$AO^2 + 9BG^2 + 4CS^2 = 0 + 8 + 5 = 13$.

(B) The circumcentre of a triangle is equidistant from its vertices. Let $O(x, y)$ be the circumcentre of the triangle with vertices $A(-2, 3), B(1, -2)$,and $C(2, 1)$.
Since $OA = OB = OC$,we have $OA^2 = OB^2 = OC^2$.
$OA^2 = (x + 2)^2 + (y - 3)^2 = x^2 + 4x + 4 + y^2 - 6y + 9 = x^2 + y^2 + 4x - 6y + 13$
$OB^2 = (x - 1)^2 + (y + 2)^2 = x^2 - 2x + 1 + y^2 + 4y + 4 = x^2 + y^2 - 2x + 4y + 5$
$OC^2 = (x - 2)^2 + (y - 1)^2 = x^2 - 4x + 4 + y^2 - 2y + 1 = x^2 + y^2 - 4x - 2y + 5$
Equating $OA^2 = OB^2$:
$x^2 + y^2 + 4x - 6y + 13 = x^2 + y^2 - 2x + 4y + 5$
$6x - 10y + 8 = 0 \Rightarrow 3x - 5y + 4 = 0 \dots (i)$
Equating $OB^2 = OC^2$:
$x^2 + y^2 - 2x + 4y + 5 = x^2 + y^2 - 4x - 2y + 5$
$2x + 6y = 0$ $\Rightarrow x + 3y = 0$ $\Rightarrow x = -3y \dots (ii)$
Substituting $(ii)$ into $(i)$:
$3(-3y) - 5y + 4 = 0$
$-9y - 5y + 4 = 0$ $\Rightarrow -14y = -4$ $\Rightarrow y = \frac{4}{14} = \frac{2}{7}$
Now,$x = -3\left(\frac{2}{7}\right) = -\frac{6}{7}$.
Thus,the circumcentre is $\left(-\frac{6}{7}, \frac{2}{7}\right)$.

(A) The vertices of the triangle are the intersection points of the lines:
$1$. $y=\sqrt{3}x$ and $y=-\sqrt{3}x$ intersect at $A(0, 0)$.
$2$. $y=\sqrt{3}x$ and $y=3$ intersect at $B(\sqrt{3}, 3)$.
$3$. $y=-\sqrt{3}x$ and $y=3$ intersect at $C(-\sqrt{3}, 3)$.
The lengths of the sides are:
$c = AB = \sqrt{(\sqrt{3}-0)^2 + (3-0)^2} = \sqrt{3+9} = \sqrt{12} = 2\sqrt{3}$.
$b = AC = \sqrt{(-\sqrt{3}-0)^2 + (3-0)^2} = \sqrt{3+9} = 2\sqrt{3}$.
$a = BC = \sqrt{(\sqrt{3}-(-\sqrt{3}))^2 + (3-3)^2} = \sqrt{(2\sqrt{3})^2} = 2\sqrt{3}$.
Since $a=b=c$,the triangle is equilateral.
The incentre $(I)$ of an equilateral triangle is the same as its centroid $(G)$.
$I = G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{0+\sqrt{3}-\sqrt{3}}{3}, \frac{0+3+3}{3}\right) = \left(0, \frac{6}{3}\right) = (0, 2)$.

(A) The vertices of the triangle are $A(a, b), B(a, c)$ and $C(d, c)$.
Since the $x$-coordinate of $A$ and $B$ is $a$,the side $AB$ is vertical. Since the $y$-coordinate of $B$ and $C$ is $c$,the side $BC$ is horizontal. Thus,$\triangle ABC$ is a right-angled triangle at vertex $B(a, c)$.
The orthocentre of a right-angled triangle is the vertex at which the right angle is formed. Therefore,the orthocentre is $B(a, c)$.
The centroid $G$ of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
$G = \left(\frac{a+a+d}{3}, \frac{b+c+c}{3}\right) = \left(\frac{2 a+d}{3}, \frac{b+2 c}{3}\right)$.
The mid-point of the line segment joining the centroid $G\left(\frac{2 a+d}{3}, \frac{b+2 c}{3}\right)$ and the orthocentre $H(a, c)$ is:
$M = \left(\frac{\frac{2 a+d}{3}+a}{2}, \frac{\frac{b+2 c}{3}+c}{2}\right) = \left(\frac{2 a+d+3 a}{6}, \frac{b+2 c+3 c}{6}\right) = \left(\frac{5 a+d}{6}, \frac{b+5 c}{6}\right)$.

(A) Given lines are:
$L_1: x+y+1=0$
$L_2: x-y-1=0$
$L_3: 3x+4y+5=0$
Slope of $L_1$ $(m_1)$ is $-1$.
Slope of $L_2$ $(m_2)$ is $1$.
Since $m_1 \times m_2 = (-1) \times (1) = -1$,the lines $L_1$ and $L_2$ are perpendicular to each other.
Therefore,the triangle is a right-angled triangle,and the orthocentre of a right-angled triangle is the vertex where the right angle is formed.
To find the vertex,solve $L_1$ and $L_2$:
$x+y+1=0$
$x-y-1=0$
Adding the two equations: $2x = 0 \Rightarrow x = 0$.
Substituting $x=0$ in $x+y+1=0$,we get $0+y+1=0 \Rightarrow y = -1$.
Thus,the orthocentre is $(0, -1)$.

(B) Given vertices are $A(0,0)$,$B(0,2)$,and $C(2,0)$.
Since $\overline{AC}$ lies on the $x$-axis and $\overline{AB}$ lies on the $y$-axis,the triangle is a right-angled triangle with the right angle at vertex $A(0,0)$.
In a right-angled triangle,the orthocentre is the vertex where the right angle is formed,so the orthocentre is $H(0,0)$.
The circumcentre of a right-angled triangle is the midpoint of the hypotenuse $\overline{BC}$.
Circumcentre $O = \left(\frac{0+2}{2}, \frac{2+0}{2}\right) = (1,1)$.
The distance between the orthocentre $(0,0)$ and the circumcentre $(1,1)$ is $\sqrt{(1-0)^2 + (1-0)^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \text{ units}$.

(B) Let the vertices be $A=(-1, 3)$,$B=(2, 5)$,and $C=(a, b)$. The orthocenter is $H=(1, 2)$.
Since $AH \perp BC$,the slope of $AH$ $(m_{AH})$ multiplied by the slope of $BC$ $(m_{BC})$ is $-1$.
$m_{AH} = \frac{2-3}{1-(-1)} = \frac{-1}{2}$.
$m_{BC} = \frac{b-5}{a-2}$.
Since $m_{AH} \times m_{BC} = -1$,we have $\left(\frac{-1}{2}\right) \times \left(\frac{b-5}{a-2}\right) = -1 \Rightarrow b-5 = 2(a-2) \Rightarrow b-5 = 2a-4 \Rightarrow 2a-b = -1$ ... $(i)$
Similarly,since $BH \perp AC$,the slope of $BH$ $(m_{BH})$ multiplied by the slope of $AC$ $(m_{AC})$ is $-1$.
$m_{BH} = \frac{2-5}{1-2} = \frac{-3}{-1} = 3$.
$m_{AC} = \frac{b-3}{a-(-1)} = \frac{b-3}{a+1}$.
Since $m_{BH} \times m_{AC} = -1$,we have $3 \times \left(\frac{b-3}{a+1}\right) = -1 \Rightarrow 3b-9 = -a-1 \Rightarrow a+3b = 8$ ... (ii)
Solving equations $(i)$ and (ii):
From $(i)$,$b = 2a+1$. Substituting into (ii): $a + 3(2a+1) = 8 \Rightarrow a + 6a + 3 = 8 \Rightarrow 7a = 5 \Rightarrow a = \frac{5}{7}$.
Then $b = 2(\frac{5}{7}) + 1 = \frac{10}{7} + \frac{7}{7} = \frac{17}{7}$.
Thus,$(a, b) = \left(\frac{5}{7}, \frac{17}{7}\right)$.

(C) Let the vertices be $A(-2, -1)$,$B(2, 5)$,and $C(m, n)$. Let $H\left(2, \frac{5}{3}\right)$ be the orthocenter.
Slope of $AH = \frac{\frac{5}{3} - (-1)}{2 - (-2)} = \frac{8/3}{4} = \frac{2}{3}$.
Since $AH \perp BC$,the slope of $BC = -\frac{1}{2/3} = -\frac{3}{2}$.
Equation of $BC$: $y - 5 = -\frac{3}{2}(x - 2)$ $\Rightarrow 2y - 10 = -3x + 6$ $\Rightarrow 3x + 2y = 16 \quad ...(i)$
Slope of $BH = \frac{\frac{5}{3} - 5}{2 - 2} = \frac{-10/3}{0}$,which is undefined. This means $BH$ is a vertical line $x = 2$.
Since $BH \perp AC$,$AC$ must be a horizontal line. Since $A$ is $(-2, -1)$,the equation of $AC$ is $y = -1$.
Since $C(m, n)$ lies on $AC$,$n = -1$.
Substitute $n = -1$ into equation $(i)$: $3m + 2(-1) = 16$ $\Rightarrow 3m = 18$ $\Rightarrow m = 6$.
Thus,the third vertex is $(6, -1)$.
Therefore,$m + n = 6 + (-1) = 5$.

(B) Let the lines be $L_1: x+y+10=0$,$L_2: x-y-2=0$,and $L_3: 2x+y-7=0$.
First,check the slopes of the lines:
Slope of $L_1$ is $m_1 = -1$.
Slope of $L_2$ is $m_2 = 1$.
Since $m_1 \times m_2 = (-1) \times (1) = -1$,the lines $L_1$ and $L_2$ are perpendicular to each other.
In a right-angled triangle,the orthocenter is the vertex where the right angle is formed.
Therefore,the orthocenter is the point of intersection of $L_1$ and $L_2$.
Solving $x+y+10=0$ and $x-y-2=0$:
Adding the two equations: $(x+y+10) + (x-y-2) = 0 \implies 2x + 8 = 0 \implies x = -4$.
Substituting $x = -4$ into $x-y-2=0$: $-4 - y - 2 = 0 \implies y = -6$.
Thus,the orthocenter is $(-4, -6)$.

(A) Let the intersection of lines $x-3y+3=0$ and $x+3y+3=0$ be $A$. Adding the equations: $2x+6=0 \implies x=-3$. Substituting $x=-3$ in $x-3y+3=0$,we get $y=0$. So,$A = (-3, 0)$.
Let the intersection of lines $x+3y+3=0$ and $x+y-1=0$ be $B$. Subtracting the equations: $(x+3y+3) - (x+y-1) = 0 \implies 2y+4=0 \implies y=-2$. Substituting $y=-2$ in $x+y-1=0$,we get $x-2-1=0 \implies x=3$. So,$B = (3, -2)$.
Let the intersection of lines $x-3y+3=0$ and $x+y-1=0$ be $C$. Subtracting the equations: $(x-3y+3) - (x+y-1) = 0 \implies -4y+4=0 \implies y=1$. Substituting $y=1$ in $x+y-1=0$,we get $x+1-1=0 \implies x=0$. So,$C = (0, 1)$.
The centroid $G$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
$G = \left(\frac{-3+3+0}{3}, \frac{0-2+1}{3}\right) = \left(0, -\frac{1}{3}\right)$.

(D) Let the coordinates of point $P$ be $(x, y)$. Since $P$ is equidistant from $A(-1, 3)$,$B(3, 5)$,and $C(5, 7)$,we have $PA = PB = PC$,which implies $PA^2 = PB^2 = PC^2$.
$PA^2 = (x+1)^2 + (y-3)^2 = x^2 + 2x + 1 + y^2 - 6y + 9 = x^2 + y^2 + 2x - 6y + 10$
$PB^2 = (x-3)^2 + (y-5)^2 = x^2 - 6x + 9 + y^2 - 10y + 25 = x^2 + y^2 - 6x - 10y + 34$
$PC^2 = (x-5)^2 + (y-7)^2 = x^2 - 10x + 25 + y^2 - 14y + 49 = x^2 + y^2 - 10x - 14y + 74$
Equating $PA^2 = PB^2$:
$x^2 + y^2 + 2x - 6y + 10 = x^2 + y^2 - 6x - 10y + 34$
$8x + 4y = 24 \implies 2x + y = 6$ $(i)$
Equating $PB^2 = PC^2$:
$x^2 + y^2 - 6x - 10y + 34 = x^2 + y^2 - 10x - 14y + 74$
$4x + 4y = 40 \implies x + y = 10$ $(ii)$
Subtracting $(ii)$ from $(i)$:
$(2x + y) - (x + y) = 6 - 10 \implies x = -4$
Substituting $x = -4$ into $(ii)$:
$-4 + y = 10 \implies y = 14$
Thus,$P$ is $(-4, 14)$.
$PA = \sqrt{(-4 - (-1))^2 + (14 - 3)^2} = \sqrt{(-3)^2 + (11)^2} = \sqrt{9 + 121} = \sqrt{130}$.
Therefore,option $(d)$ is correct.

(D) Let the vertices be $A(1, \sqrt{3})$,$B(0, 0)$,and $C(2, 0)$.
Calculate the side lengths:
$AB = \sqrt{(1-0)^2 + (\sqrt{3}-0)^2} = \sqrt{1 + 3} = 2$
$BC = \sqrt{(2-0)^2 + (0-0)^2} = \sqrt{4} = 2$
$AC = \sqrt{(2-1)^2 + (0-\sqrt{3})^2} = \sqrt{1 + 3} = 2$
Since all sides are equal,the triangle is an equilateral triangle.
For an equilateral triangle,the incentre is the same as the centroid.
The centroid $(G)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
$G = \left(\frac{1+0+2}{3}, \frac{\sqrt{3}+0+0}{3}\right) = \left(\frac{3}{3}, \frac{\sqrt{3}}{3}\right) = \left(1, \frac{1}{\sqrt{3}}\right)$.

(B) The vertices are $A(1,1), B(1,-1), C(-1,1)$.
Since $AB$ is vertical (along $x=1$) and $AC$ is horizontal (along $y=1$),$\triangle ABC$ is a right-angled triangle at $A(1,1)$.
$1$. Circumcentre $S$: For a right-angled triangle,the circumcentre is the midpoint of the hypotenuse $BC$.
$S = \left(\frac{1+(-1)}{2}, \frac{-1+1}{2}\right) = (0,0)$.
$2$. Orthocentre $O$: For a right-angled triangle,the orthocentre is the vertex at the right angle.
$O = A = (1,1)$.
$3$. Incentre $I$: The side lengths are $c = AB = \sqrt{(1-1)^2 + (1-(-1))^2} = 2$,$b = AC = \sqrt{(1-(-1))^2 + (1-1)^2} = 2$,and $a = BC = \sqrt{(1-(-1))^2 + (-1-1)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$.
The coordinates of the incentre $I$ are given by $\left(\frac{ax_A + bx_B + cx_C}{a+b+c}, \frac{ay_A + by_B + cy_C}{a+b+c}\right)$.
$I = \left(\frac{2\sqrt{2}(1) + 2(1) + 2(-1)}{2\sqrt{2}+2+2}, \frac{2\sqrt{2}(1) + 2(-1) + 2(1)}{2\sqrt{2}+2+2}\right) = \left(\frac{2\sqrt{2}}{2\sqrt{2}+4}, \frac{2\sqrt{2}}{2\sqrt{2}+4}\right) = \left(\frac{\sqrt{2}}{\sqrt{2}+2}, \frac{\sqrt{2}}{\sqrt{2}+2}\right) = \left(\frac{1}{1+\sqrt{2}}, \frac{1}{1+\sqrt{2}}\right) = (\sqrt{2}-1, \sqrt{2}-1)$.
Now,calculate $IS + OS$:
$OS = \sqrt{(1-0)^2 + (1-0)^2} = \sqrt{1+1} = \sqrt{2}$.
$IS = \sqrt{(\sqrt{2}-1-0)^2 + (\sqrt{2}-1-0)^2} = \sqrt{2(\sqrt{2}-1)^2} = \sqrt{2}(\sqrt{2}-1) = 2 - \sqrt{2}$.
$IS + OS = (2 - \sqrt{2}) + \sqrt{2} = 2$.

(C) Let the vertices be $A(1, 2)$,$B(3, -1)$,and $C(4, 0)$.
First,calculate the lengths of the sides:
$AB = \sqrt{(3-1)^2 + (-1-2)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4+9} = \sqrt{13}$
$AC = \sqrt{(4-1)^2 + (0-2)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9+4} = \sqrt{13}$
Since $AB = AC$,the triangle is isosceles.
Centroid $G = \left(\frac{1+3+4}{3}, \frac{2-1+0}{3}\right) = \left(\frac{8}{3}, \frac{1}{3}\right)$.
For an isosceles triangle,the circumcentre $O$ lies on the altitude from $A$ to $BC$. The slope of $BC$ is $m_{BC} = \frac{0 - (-1)}{4 - 3} = 1$. The altitude from $A$ has slope $-1$ and passes through $(1, 2)$,so its equation is $y - 2 = -1(x - 1) \Rightarrow y = -x + 3$.
The circumcentre $O$ is the intersection of the perpendicular bisectors. The perpendicular bisector of $AC$ passes through the midpoint $E\left(\frac{1+4}{2}, \frac{2+0}{2}\right) = \left(\frac{2.5}{1}, 1\right)$ with slope $m = -\frac{1}{m_{AC}} = -\frac{1}{-2/3} = \frac{3}{2}$.
Equation of perpendicular bisector of $AC$: $y - 1 = \frac{3}{2}(x - 2.5)$ $\Rightarrow y = 1.5x - 3.75 + 1$ $\Rightarrow y = 1.5x - 2.75$.
Solving $y = -x + 3$ and $y = 1.5x - 2.75$: $-x + 3 = 1.5x - 2.75$ $\Rightarrow 2.5x = 5.75$ $\Rightarrow x = 2.3 = \frac{23}{10}$.
Then $y = -2.3 + 3 = 0.7 = \frac{7}{10}$. So $O = \left(\frac{23}{10}, \frac{7}{10}\right)$.
The distance $OG = \sqrt{(\frac{8}{3} - \frac{23}{10})^2 + (\frac{1}{3} - \frac{7}{10})^2} = \sqrt{(\frac{80-69}{30})^2 + (\frac{10-21}{30})^2} = \sqrt{(\frac{11}{30})^2 + (-\frac{11}{30})^2} = \frac{11}{30} \sqrt{2}$.

(D) The intersection of the perpendicular bisectors of the sides of a triangle is the circumcenter $O$. Solving the equations $x-y+5=0$ and $x+2y=0$,we get $x = -10/3$ and $y = 5/3$. Thus,the circumcenter $O$ is $(-10/3, 5/3)$.
Since $B$ is the reflection of $A(1,-2)$ across the line $x-y+5=0$,we have $\frac{x_B-1}{1} = \frac{y_B+2}{-1} = -2 \frac{1-(-2)+5}{1^2+(-1)^2} = -2 \frac{8}{2} = -8$. This gives $x_B = -7$ and $y_B = 6$. So $B = (-7, 6)$.
Since $C$ is the reflection of $A(1,-2)$ across the line $x+2y=0$,we have $\frac{x_C-1}{1} = \frac{y_C+2}{2} = -2 \frac{1+2(-2)}{1^2+2^2} = -2 \frac{-3}{5} = 6/5$. This gives $x_C = 1+6/5 = 11/5$ and $y_C = -2+12/5 = 2/5$. So $C = (11/5, 2/5)$.
The perpendicular bisector of $BC$ passes through the circumcenter $O(-10/3, 5/3)$ and the midpoint $M$ of $BC$. The midpoint $M$ is $(\frac{-7+11/5}{2}, \frac{6+2/5}{2}) = (-12/5, 16/5)$.
The slope of the line passing through $O(-10/3, 5/3)$ and $M(-12/5, 16/5)$ is $m = \frac{16/5 - 5/3}{-12/5 - (-10/3)} = \frac{48/15 - 25/15}{-36/15 + 50/15} = \frac{23/15}{14/15} = 23/14$.
The equation of the line is $y - 5/3 = 23/14(x + 10/3)$,which simplifies to $14(3y-5) = 69(x+10/3)$ $\Rightarrow 42y - 70 = 69x + 230$ $\Rightarrow 69x - 42y + 300 = 0$ $\Rightarrow 23x - 14y + 100 = 0$.