Centres of a Triangle
Centroid
Consider a triangle formed by a pair of straight lines passing through the origin, \(ax^2 + 2hxy + by^2 = 0\), and a line \(lx + my + n = 0\). Then the centroid \((\alpha, \beta)\) of this triangle is given by:
\[
\frac{\alpha}{bl - hm} = \frac{\beta}{am - hl} = \frac{-2n}{3(am^2 - 2hlm + bl^2)}
\]
Proof:
Let \(y = m_1 x\) and \(y = m_2 x\) be lines represented by \(ax^2 + 2hxy + by^2 = 0\). Then,
\[
m_1 + m_2 = -\frac{2h}{b} \quad \text{and} \quad m_1 m_2 = \frac{a}{b}
\]
Let the line \(lx + my + n = 0\) cut the pair at points \(A\) and \(B\). We wish to calculate the centroid of \(\triangle OAB\). To do that, we find \(A\) and \(B\).
\(A\) is the intersection point of \(y = m_1 x\) and \(lx + my + n = 0\): Eliminating y,
\[
lx + mm_1x + n = 0 \implies x = \frac{-n}{l + mm_1}
\]
\[
\implies A = \left( \frac{-n}{l + mm_1}, \frac{-m_1 n}{l + mm_1} \right)
\]
Similarly, \(B\) is the intersection point of \(y = m_2 x\) and \(lx + m y + n = 0\):
\[
\implies B = \left( \frac{-n}{l + mm_2}, \frac{-m_2 n}{l + mm_2} \right)
\]
The centroid of \(\triangle OAB\) is:
\[
\left( \frac{0 + x_A + x_B}{3}, \frac{0 + y_A + y_B}{3} \right)
\]
For the \(x\)-coordinate:
\[
\frac{x_A + x_B}{3} = \frac{-n}{3} \left( \frac{1}{l + mm_1} + \frac{1}{l + mm_2} \right)
\]
\[
= \frac{-n}{3} \left( \frac{(l + mm_2) + (l + mm_1)}{(l + mm_1)(l + mm_2)} \right)
\]
\[
= \frac{-n}{3} \left( \frac{2l + m(m_1 + m_2)}{l^2 + m l (m_1 + m_2) + m^2 m_1 m_2} \right)
\]
\[
= \frac{-n}{3} \left( \frac{2l - m \frac{2h}{b}}{l^2 - \frac{2hlm}{b} + \frac{am^2}{b}} \right)
\]
\[
= \frac{-n}{3} \left( \frac{2bl - 2hm}{bl^2 - 2hlm + am^2} \right)
\]
\[
= \frac{-2n (bl - hm)}{3(bl^2 - 2hlm + am^2)}
\]
For the \(y\)-coordinate:
\[
\frac{y_A + y_B}{3} = \frac{-n m_1}{3 (l + mm_1)} + \frac{-n m_2}{3 (l + mm_2)}
\]
\[
= \frac{-n}{3} \left( \frac{m_1 (l + mm_2) + m_2 (l + mm_1)}{(l + mm_1) (l + mm_2)} \right)
\]
\[
= \frac{-n}{3} \left( \frac{m_1 l + m_1^2 m + m_2 l + m_2^2 m}{l^2 + lm (m_1 + m_2) + m^2 m_1 m_2} \right)
\]
\[
= \frac{-n}{3} \left( \frac{l (m_1 + m_2) + 2m m_1 m_2}{l^2 + lm (m_1 + m_2) + m^2 m_1 m_2} \right)
\]
\[
= \frac{-n}{3} \left( \frac{l \frac{-2h}{b} + 2m \frac{a}{b}}{l^2 - lm \frac{2h}{b} + m^2 \frac{a}{b}} \right)
\]
\[
= \frac{-2n (hl - am)}{3 (bl^2 - 2hlm + am^2)}
\]
So, the centroid \((\alpha, \beta)\) is given by:
\[
\alpha = \frac{-2n (bl - hm)}{3 (bl^2 - 2hlm + am^2)} \quad \text{and} \quad \beta = \frac{-2n (hl - am)}{3 (bl^2 - 2hlm + am^2)}
\]
\[
\implies \frac{\alpha}{bl - hm} = \frac{\beta}{am - hl} = \frac{-2n}{3 (am^2 - 2hlm + bl^2)}
\]
Orthocentre
Consider a pair of straight lines \( ax^2 + 2hxy + by^2 = 0 \) forming a triangle with \( lx + my + n = 0 \). We wish to find the orthocentre of this triangle.
To do that we first finf the points of intersection \(A\) and \(B\) of the line with the pair of straight lines. we have already done so in the calculation of centroid. So for the sake of efficiency let's use that.
Coordinates of point \( A \) are:
\[
\left( \frac{-n}{l + mm_1}, \frac{-nm_1}{l + mm_1} \right)
\]
Coordinates of point \( B \) are:
\[
\left( \frac{-n}{l + mm_2}, \frac{-nm_2}{l + mm_2} \right)
\]
We know that the orthocenter \( H (\alpha, \beta) \) is the intersection of altitudes of the \( \Delta OAB \). Consider altitudes \( AM \) and \( BN \).
\[
-\frac{1}{m_2}
\]
\[
-\frac{1}{m_1}
\]
\[
-\frac{1}{m_2} = \frac{y - y_A}{x - x_A} \implies y = -\frac{1}{m_2}(x - x_A) + y_A \tag{1}
\]
\[
-\frac{1}{m_1} = \frac{y - y_B}{x - x_B} \implies y = -\frac{1}{m_1}(x - x_B) + y_B \tag{2}
\]
Orthocenter \( H (\alpha, \beta) \) is the intersection point of \( (1) \) and \( (2) \). Eliminating \( y \):
\[\begin{align*}
-\frac{1}{m_2}(x - x_A) + y_A &= -\frac{1}{m_1}(x - x_B) + y_B \\
\left( \frac{1}{m_1} - \frac{1}{m_2} \right) x &= \frac{x_B}{m_1} - \frac{x_A}{m_2} + y_B - y_A \\
\end{align*}\]
\[
\left( \frac{1}{m_1} - \frac{1}{m_2} \right) x = \frac{x_B}{m_1} - \frac{x_A}{m_2} + y_B - y_A
\]
\[
\frac{m_2 - m_1}{m_1 m_2} x = - \frac{n}{(l + mm_2) m_1} + \frac{n}{(l + mm_1) m_2} - \frac{nm_2}{(l + mm_2)} + \frac{nm_1}{(l + mm_1)}
\]
\[
\frac{m_2 - m_1}{m_1 m_2} x = - \frac{n}{l + mm_2} \left( \frac{1}{m_1} + m_2 \right) + \frac{n}{l + mm_1} \left( \frac{1}{m_2} + m_1 \right)
\]
\[
\frac{m_2 - m_1}{m_1 m_2} x = - \frac{n(1 + m_1 m_2)}{m_1 (l + m m_2)} + \frac{n(1 + m_1 m_2)}{m_2 (l + m m_1)}
\]
\[
\frac{m_2 - m_1}{m_1 m_2} x = - n (1 + m_1 m_2) \left[ \frac{m_2 l + m m_1 m_2 - m_1 l - m m_1 m_2}{m_1 m_2 (l + m m_1) (l + m m_2)} \right]
\]
\[
\frac{m_2 - m_1}{m_1 m_2} x = - n (1 + m_1 m_2) \left[ \frac{l (m_2 - m_1)}{m_1 m_2 (l^2 + lm (m_1 + m_2) + m^2 m_1 m_2)} \right]
\]
\[
x = \frac{-nl \left(1 + m_1 m_2 \right)}{l^2 + lm \left(m_1 + m_2 \right) + m^2 m_1 m_2}
= \frac{-nl \left(1 + \frac{a}{b} \right)}{l^2 - \frac{2mh}{b} + \frac{m^2 a}{b}}
= \frac{-nl \left(a + b \right)}{bl^2 - 2hlm + am^2}
\]
Thus:
\[
\alpha = -\frac{nl (a + b)}{bl^2 - 2hlm + am^2}
\]
Since \( (\alpha, \beta) \) also lies on the third altitude from origin whose slope is \( \frac{m}{l} \):
\[\begin{align*}
\beta &= \frac{m}{l} \alpha \\
&\implies \beta = \frac{m}{l} \left( -\frac{nl (a + b)}{bl^2 - 2hlm + am^2} \right) \\
&\implies \beta = -\frac{nm (a + b)}{bl^2 - 2hlm + am^2}
\end{align*}\]
We can write this as:
\[
\frac{\alpha}{l} = \frac{\beta}{m} = -\frac{n(a + b)}{bl^2 - 2hlm + am^2}
\]
Equation of a line with known midpoint
Consider the pair of straight lines given by the equation \(ax^2 + 2hxy + by^2 = 0\) and a line \(L\) that intersects these lines at points \(A\) and \(B\). Suppose the midpoint of the segment \(AB\) is known to be \((x_1, y_1)\). We aim to determine the equation of the line \(L\).
To find the equation of line \(L\), we only need to find it's slope because we already know a point on it. Let the slope of \(L\) be \(m\). Thus, the equation of \(L\) can be expressed in slope-intercept form as:
\[
y = mx + c
\]
Since \((x_1, y_1)\) lies on \(L\), we substitute these coordinates into the line equation:
\[
y_1 = mx_1 + c \implies c = y_1 - mx_1
\]
Thus, the equation of \(L\) becomes:
\[
y = mx + (y_1 - mx_1)
\]
Next, we substitute \(y = mx + (y_1 - mx_1)\) into the equation of the pair of lines \(ax^2 + 2hxy + by^2 = 0\) to find the points of intersection \(A\) and \(B\):
\[
ax^2 + 2h x (mx + y_1 - mx_1) + b (mx + y_1 - mx_1)^2 = 0
\]
This simplifies to a quadratic equation in \(x\), which can be written as:
\[
ax^2 + 2hx(mx + y_1 - mx_1) + b(mx + y_1 - mx_1)^2 = 0
\]
Expanding and simplifying, we find:
\[
ax^2 + 2h x (mx + y_1 - mx_1) + b(m^2x^2 + 2m x (y_1 - mx_1) + (y_1 - mx_1)^2) = 0
\]
Grouping like terms, the quadratic equation in \(x\) becomes:
\[
(a + 2hm + bm^2)x^2 + 2(h + bm)(y_1 - mx_1)x + b(y_1 - mx_1)^2 = 0
\]
Absciisas of \(A\) and \(B\), that is, \(x_A\) and \(x_B\) are the roots of this quadratic equation. Since \((x_1, y_1)\) is the midpoint of \(A(x_A, y_A)\) and \(B(x_B, y_B)\), we have:
\[
\frac{x_A + x_B}{2} = x_1
\]
From the properties of the roots of a quadratic equation, the sum of the roots \(x_A\) and \(x_B\) is given by:
\[
x_A + x_B = - \frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} = - \frac{2(h + bm)(y_1 - mx_1)}{a + 2hm + bm^2}
\]
Thus,
\[
2x_1 = - \frac{2(h + bm)(y_1 - mx_1)}{a + 2hm + bm^2} \implies x_1 = -\frac{(h + bm)(y_1 - mx_1)}{a + 2hm + bm^2}
\]
Rewriting this, we obtain:
\[
(a + 2hm + bm^2)x_1 = -(h + bm)(y_1 - mx_1)
\]
\[
a x_1 + 2hmx_1 + bm^2 x_1 = -h y_1 + h mx_1 - b y_1 m + bm^2 x_1
\]
\[
a x_1 + h mx_1 + b m y_1 + h y_1 = 0
\]
Solving for \(m\):
\[
m = -\frac{ax_1 + h y_1}{h x_1 + b y_1}
\]
Thus, the slope \(m\) is found. Now, substituting back, the equation of the line \(L\) in point-slope form is:
\[
\frac{-(ax_1 + h y_1)}{h x_1 + b y_1} = \frac{y - y_1}{x - x_1}
\]
Expanding and verifying, we obtain:
\[
-ax_1 + ax_1^2 - hy_1x + hy_1x_1 = hx_1y - hx_1y_1 + byy_1 - by^2
\]
\[
ax_1^2 + by^2 + 2hx_1y = h(xy_1 + yx_1) + ax_1 + byy_1
\]
\[
\boxed{ax_1^2 + 2hx_1y + by^2 = ax_1 + h(xy_1 + yx_1) + byy_1}\]
Hence, we have derived the equation of the line \(L\) that intersects the given pair of lines at points \(A\) and \(B\), with the midpoint of the segment \(AB\) being \((x_1, y_1)\).