Triangles and Pair of Straight Lines
Triangles formed by a pair of lines with another line
Consider a pair of straight lines \( ax^2 + 2hxy + by^2 = 0 \) and a line \( lx + my + n = 0 \). They intersect at two points, \( A \) and \( B \). We are interested in the nature of \( \Delta OAB \).
The triangle \( \Delta OAB \) formed by the lines represented by \( ax^2 + 2hxy + by^2 = 0 \) and a line \( lx + my + n = 0 \) is isosceles if \( h(l^2  m^2) = (a  b)lm \).
Proof:
Let \( y = m_1x \) and \( y = m_2x \) be lines represented by \( ax^2 + 2hxy + by^2 = 0 \).
Slope of line \( AB \) is \(\frac{l}{m} \).
Since \( \angle OAB = \angle OBA \):
Substituting \( m_1 + m_2 = \frac{2h}{b} \) and \( m_1m_2 = \frac{a}{b} \):
Therefore, the triangle \( \Delta OAB \) is isosceles if:
This completes the proof.
Consider a straight line \( ax + by + c = 0 \). We wish to find a pair of straight lines passing through the origin and forming an isosceles triangle with \( ax + by + c = 0 \) as the base, where the vertex angle is \( \alpha \).
Theorem: The equation of a pair of straight lines passing through the origin and forming an isosceles triangle with a given line \( ax + by + c = 0 \) as the base, where the vertex angle is \( \alpha \), is \( (ax + by)^2  \tan^2 \alpha (bx  ay)^2 = 0 \).
Proof:
Let the lines \( OA \) and \( OB \) be represented as shown in the figure. Let \((x, y)\) be a point on \( OA \). The slope of \( OA \) is \( \frac{y}{x} \).
The slope of the line \( ax + by + c = 0 \) is \( \frac{a}{b} \).
Given that \( \angle OAB = \alpha \):
Therefore, the equation of the pair of lines is:
This completes the proof.
Special Cases:

If \( \alpha = 60^\circ \):
 The triangle formed is equilateral.
 The equation becomes:
\[ (ax + by)^2  3(bx  ay)^2 = 0 \] 
If \( \alpha = 45^\circ \):
 The triangle formed is a rightangled isosceles triangle.
 The equation becomes:
\[ (ax + by)^2  (bx  ay)^2 = 0 \]
A pair of straight lines \( ax^2 + 2hxy + by^2 = 0 \) forms an equilateral triangle with the line \( lx + my + n = 0 \) if it can be expressed as:
e.g. The pair of straight lines \(3x^2 + 48xy + 23y^2 = 0\), and the line is \(3x  2y + 4 = 0\) form an equilateral triangleWe can show this by expressing it as \((3x  2y)^2  3(2x + 3y)^2 = 0\). Observe that
Thus,
which simplifies to:
Therefore, the pair of straight lines \(3x^2 + 48xy + 23y^2 = 0\) and the line \(3x  2y + 4 = 0\) form an equilateral triangle because the given pair of lines can be expressed as \((3x  2y)^2  3(2x + 3y)^2 = 0\).
A line \(lx + my + n = 0\) forms a rightangled triangle with \(ax^2 + 2hxy + by^2 = 0\) if \(a + b = 0\) or \(al^2 + 2hlm + bm^2 = 0\).
Proof:
We know that the angle between a pair of lines is a right angle if \(a + b = 0\). Thus, whatever the line \(lx + my + n = 0\), if it is not parallel to a line in the pair, then we always get a rightangled triangle.
A rightangled triangle can also be obtained if the line \(lx + my + n = 0\) is perpendicular to one of the lines of \(ax^2 + 2hxy + by^2 = 0\).
Let \(y = m_1x\) and \(y = m_2x\) be lines represented by \(ax^2 + 2hxy + by^2 = 0\).
Let \(lx + my + n = 0\) be perpendicular to \(y = m_1x\).
Then:
We know that:
Hence, a line \(lx + my + n = 0\) forms a rightangled triangle with \(ax^2 + 2hxy + by^2 = 0\) if:
Area of Triangles formed by a pair of straight lines with a given straight line
Consider the triangle formed by a pair of straight lines \( ax^2 + 2hxy + by^2 = 0 \) and the line \( lx + my + n = 0 \). We wish to find the area of this triangle.
Let \( y = m_1x \) and \( y = m_2x \) be the lines represented by \( ax^2 + 2hxy + by^2 = 0 \).
Let \( A \) and \( B \) be the intersection points of \( lx + my + n = 0 \) with \( y = m_1x \) and \( y = m_2x \), respectively. First, we find \( A \) and \( B \).
For \( A \):
Therefore, \( A \) is:
Similarly, for \( B \):
The area of \( \Delta OAB \) is given by:
Substituting the coordinates:
Therefore, the area of the triangle is:
Area of isosceles triangle formed by a line with a pair of striaght lines
Consider a straight line \( ax + by + c = 0 \) and a pair of straight lines passing through the origin, forming an isosceles triangle with the base as \( ax + by + c = 0 \) and base angle \(\alpha\). We know that the equation of the pair of straight lines is:
We are now interested in finding the area of this isosceles triangle.
First, we need to calculate the altitude \( ON \) of \( \Delta OAB \). This is the perpendicular distance from the origin \( O \) to the line \( AB \), which is the base of the triangle.
Next, we calculate \( AN \), which is the segment of the altitude from point \( A \) (one end of the base) to the point \( N \) (where the altitude intersects the base). Given that \( \angle OAN = \alpha \), we use the relationship between the altitude and the base angle:
Since the triangle is isosceles, the base \( AB \) can be expressed as twice the length of \( AN \):
Now, we can calculate the area of \( \Delta OAB \). The area of a triangle is given by half the product of its base and height:
Substituting the expressions for \( AB \) and \( ON \):
Substituting the value of \( ON \):
Thus:
Therefore, the area of the isosceles triangle formed by the pair of straight lines \( (ax + by)^2  \tan^2 \alpha (bx  ay)^2 = 0 \) and the line \( ax + by + c = 0 \) is: