Q.Find the orthocenter of the triangle whose vertices are $(-2, -1)$, $(6, -1)$ and $(2, 5)$.

Two altitudes meet at $\left(2,\tfrac53\right)$, the orthocenter.

Let $A(-2,-1)$, $B(6,-1)$, $C(2,5)$. Since $A$ and $B$ share $y=-1$, side $AB$ is horizontal, so the altitude from $C$ is vertical: $x = 2$.

Altitude from $A$, perpendicular to $BC$: slope of $BC = \dfrac{5-(-1)}{2-6} = \dfrac{6}{-4} = -\dfrac32$, so the altitude slope is $\dfrac23$. Through $A(-2,-1)$:

$y + 1 = \frac23(x + 2).$

Substitute $x = 2$ (from the first altitude): $y + 1 = \dfrac23(4) = \dfrac83 \Rightarrow y = \dfrac83 - 1 = \dfrac53$.

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Find the equations of two altitudes (each perpendicular to the opposite side) and solve them simultaneously.

Let $A=(5,-2),\ B=(-1,2),\ C=(1,4)$.

Altitude from $A$ (perpendicular to $BC$):

$\text{slope of }BC = \frac{4-2}{1-(-1)} = \frac{2}{2} = 1$

So the altitude from $A$ has slope $-1$:

$y-(-2) = -1(x-5) \implies x+y = 3 \quad \text{...(1)}$

Altitude from $B$ (perpendicular to $AC$):

$\text{slope of }AC = \frac{4-(-2)}{1-5} = \frac{6}{-4} = -\frac{3}{2}$

So the altitude from $B$ has slope $\frac{2}{3}$:

$y-2 = \frac{2}{3}(x+1) \implies 2x-3y = -8 \quad \text{...(2)}$

Solve (1) and (2): From (1), $x = 3-y$. Substitute into (2): …

The circumcenter is equidistant from all three vertices; setting up and solving two equations (pairwise equal distances) gives its coordinates.

Let the circumcenter be $S(x,y)$, equidistant from $A(1,3)$, $B(-3,5)$, $C(5,-1)$.

$SA^2=SB^2$:

$(x-1)^2+(y-3)^2 = (x+3)^2+(y-5)^2$

$x^2-2x+1+y^2-6y+9 = x^2+6x+9+y^2-10y+25$

$-2x-6y+10 = 6x-10y+34$

$-8x+4y-24=0 \Rightarrow 2x-y+6=0 \quad \text{...(i)}$

$SA^2=SC^2$:

$(x-1)^2+(y-3)^2 = (x-5)^2+(y+1)^2$

$x^2-2x+1+y^2-6y+9 = x^2-10x+25+y^2+2y+1$

$-2x-6y+10 = -10x+2y+26$

$8x-8y-16=0 \Rightarrow x-y-2=0 \quad \text{...(ii)}$

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Each side of the triangle lies along one of the three given lines. Find the vertices as pairwise intersections, then intersect two altitudes (each perpendicular to the opposite side).

Let $L_1: x+2y=0$, $L_2: 4x+3y-5=0$, $L_3: 3x+y=0$.

Vertices:

$A=L_1\cap L_2$: from $x=-2y$, $4(-2y)+3y-5=0 \Rightarrow -5y=5 \Rightarrow y=-1,\ x=2$, so $A(2,-1)$.

$B=L_1\cap L_3$: from $x=-2y$ and $y=-3x$, solving gives $x=0,y=0$, so $B(0,0)$.

$C=L_2\cap L_3$: with $y=-3x$, $4x+3(-3x)-5=0 \Rightarrow -5x=5 \Rightarrow x=-1,y=3$, so $C(-1,3)$.

Since $B,C$ both lie on $L_3$, side $BC$ is along $L_3$ ($3x+y=0$, slope $-3$); since $A,C$ both lie on $L_2$, side $AC$ is along $L_2$ (slope $-4/3$).

Altitude from $A$, perpendicular to $BC$ (slope $-3$, so altitude slope $=1/3$) through $A(2,-1)$:

$y+1=\dfrac13(x-2) \Rightarrow x-3y-5=0$

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$A(-2,-1)$ and $B(6,-1)$ share $y=-1$, so the altitude from $C(2,5)$ is $x=2$; intersect it with the altitude from $A$ to get $\left(2,\frac53\right)$.

Let $A(-2,-1)$, $B(6,-1)$, $C(2,5)$.

Since $A$ and $B$ have the same $y$-coordinate, side $AB$ is horizontal, so the altitude from $C$ (perpendicular to $AB$) is the vertical line $x=2$.

Altitude from $A$ is perpendicular to $BC$. Slope of $BC=\dfrac{5-(-1)}{2-6}=\dfrac{6}{-4}=-\dfrac{3}{2}$, so the altitude from $A$ has slope $\dfrac{2}{3}$:

$y+1=\dfrac{2}{3}(x+2)$.

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Find two altitudes (each perpendicular to a side, through the opposite vertex) and solve them simultaneously.

Let $A(5,-2)$, $B(-1,2)$, $C(1,4)$.

Altitude from $A$, perpendicular to $BC$:

Slope of $BC = \dfrac{4-2}{1-(-1)} = \dfrac{2}{2}=1$. Perpendicular slope $=-1$.

Line through $A(5,-2)$ with slope $-1$:

$y+2 = -1(x-5) \implies x+y=3 \quad \text{...(1)}$

Altitude from $B$, perpendicular to $AC$:

Slope of $AC = \dfrac{4-(-2)}{1-5} = \dfrac{6}{-4}=-\dfrac{3}{2}$. Perpendicular slope $=\dfrac{2}{3}$.

Line through $B(-1,2)$ with slope $\dfrac{2}{3}$:

$y-2=\frac{2}{3}(x+1) \implies 3y-6=2x+2 \implies 2x-3y=-8 \quad \text{...(2)}$

Solving (1) and (2): From (1), $x=3-y$. Substitute into (2):

$2(3-y)-3y=-8 \implies 6-2y-3y=-8 \implies -5y=-14 \implies y=\frac{14}{5}$

$x = 3-\frac{14}{5} = \frac{1}{5}$

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Solving the equidistance conditions gives circumcentre $\left(-\tfrac13,\tfrac23\right)$.

Let the circumcentre be $S=(x,y)$, equidistant from $A(1,3),\ B(0,-2),\ C(-3,1)$.

$SA^2=SB^2$:

$(x-1)^2+(y-3)^2=x^2+(y+2)^2.$

$-2x+1-6y+9=4y+4\implies -2x-10y+6=0\implies x+5y=3.\quad(i)$

$SB^2=SC^2$:

$x^2+(y+2)^2=(x+3)^2+(y-1)^2.$

$4y+4=6x-2y+10\implies -6x+6y=6\implies y=x+1.\quad(ii)$

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