Triangles and Quadrilaterals formed by lines

Ratio of sides when equations of sides are given

Consider three lines:

\[ L_1: a_1x + b_1y + c_1 = 0 \]

\[ L_2: a_2x + b_2y + c_2 = 0 \]

\[ L_3: a_3x + b_3y + c_3 = 0 \]

\(L_2\) and \(L_3\) intersect at point \(A\), \(L_3\) and \(L_1\) intersect at point \(B\), and \(L_1\) and \(L_2\) intersect at point \(C\).

Coordinates of point \(A\) are given by:

\[ x_A = \frac{\begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix}}{\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}}, \quad y_A = -\frac{\begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix}}{\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}} \]

Similarly, coordinates of \(B\) are:

\[ \left( \frac{\begin{vmatrix} b_3 & c_3 \\ b_1 & c_1 \end{vmatrix}}{\begin{vmatrix} a_3 & b_3 \\ a_1 & b_1 \end{vmatrix}}, -\frac{\begin{vmatrix} a_3 & c_3 \\ a_1 & c_1 \end{vmatrix}}{\begin{vmatrix} a_3 & b_3 \\ a_1 & b_1 \end{vmatrix}} \right) \]

Coordinates of \(C\) are:

\[ \left( \frac{\begin{vmatrix} b_1 & c_1 \\ b_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}}, -\frac{\begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}} \right) \]

The distance of \(A\) from \(BC\), \( h_A \):

\[ h_A = \frac{|a_1 x_A + b_1 y_A + c_1|}{\sqrt{a_1^2 + b_1^2}} \]

\[ a_1 x_A + b_1 y_A + c_1 = a_1 \frac{\begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix}}{\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}} - b_1 \frac{\begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix}}{\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}} + c_1 \]

\[ = \frac{a_1 \begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} - b_1 \begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix} + c_1 \begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}}{\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}} \]

\[ = \frac{\begin{vmatrix} a_1 & b_2 & c_2 \\ a_2 & b_3 & c_3 \\ a_3 & b_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}} \]

Thus,

\[ h_A = \frac{\Delta}{\sqrt{a_1^2 + b_1^2}\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}} \]

where, \(\Delta=\begin{vmatrix} a_1 & b_2 & c_2 \\ a_2 & b_3 & c_3 \\ a_3 & b_2 & c_2 \end{vmatrix}\)

Similarly,

\[ h_B = \frac{\Delta}{\sqrt{a_2^2 + b_2^2}\begin{vmatrix} a_1 & b_3 \\ a_1 & b_3 \end{vmatrix}}, \quad h_C = \frac{\Delta}{\sqrt{a_3^2 + b_3^2}\begin{vmatrix} a_1 & b_2 \\ a_1 & b_2 \end{vmatrix}} \]

Area of triangle \(ABC\), which can be written in three ways as:

\[ \text{Area} = \frac{1}{2} BC \cdot h_A = \frac{1}{2} CA \cdot h_B = \frac{1}{2} AB \cdot h_C \]

\[ \implies BC : CA : AB = \frac{1}{h_A} : \frac{1}{h_B} : \frac{1}{h_C} \]

\[ \implies BC:CA:AB = \sqrt{a_1^2 + b_1^2}\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}: \sqrt{a_2^2 + b_2^2}\begin{vmatrix} a_1 & b_3 \\ a_1 & b_3 \end{vmatrix}:\sqrt{a_3^2 + b_3^2}\begin{vmatrix} a_1 & b_2 \\ a_1 & b_2 \end{vmatrix} \]

Area of the triangle made by a line with axes

The area of the triangle formed by the line \( ax + by + c = 0 \) with the coordinate axes is \( \frac{c^2}{2|ab|} \).

Proof:

Consider the line \( ax + by + c = 0 \).

1. Finding the x-intercept:

   Set \( y = 0 \) in the equation:

   \[ ax + c = 0 \implies x = -\frac{c}{a} \]

   Therefore, the x-intercept is \( \left( -\frac{c}{a}, 0 \right) \).

2. Finding the y-intercept:

   Set \( x = 0 \) in the equation:

   \[ by + c = 0 \implies y = -\frac{c}{b} \]

   Therefore, the y-intercept is \( \left( 0, -\frac{c}{b} \right) \).

3. Area of the triangle formed by the line and the coordinate axes:

   The vertices of the triangle are \( (0, 0) \), \( \left( -\frac{c}{a}, 0 \right) \), and \( \left( 0, -\frac{c}{b} \right) \).

   The area \( A \) of a triangle with vertices at \( (0,0) \), \( (x_1, 0) \), and \( (0, y_1) \) is given by:

   \[ A = \frac{1}{2} \left| x_1 \cdot y_1 \right| \]

   Substituting \( x_1 = -\frac{c}{a} \) and \( y_1 = -\frac{c}{b} \):

   \[ A = \frac{1}{2} \left| \left( -\frac{c}{a} \right) \left( -\frac{c}{b} \right) \right| \]
   \[ A = \frac{1}{2} \left| \frac{c^2}{ab} \right| \]
   \[ A = \frac{c^2}{2|ab|} \]

   Therefore, the area of the triangle formed by the line \( ax + by + c = 0 \) with the coordinate axes is \( \frac{c^2}{2|ab|} \).

Area of Triangle Formed by Three Lines

Consider the three lines:

\[ L_1: a_1 x + b_1 y + c_1 = 0 \]

\[ L_2: a_2 x + b_2 y + c_2 = 0 \]

\[ L_3: a_3 x + b_3 y + c_3 = 0 \]

These lines form a triangle \( \Delta ABC \), where \( L_1 \) is opposite to side \( BC \), \( L_2 \) is opposite to side \( CA \), and \( L_3 \) is opposite to side \( AB \).

Coordinates of vertices \( A, B, \) and \( C \) can be calculated using the cross-multiplication method:

\[ A = \left( \frac{\left| \begin{matrix} b_2 & c_2 \\ b_3 & c_3 \end{matrix} \right|}{\left| \begin{matrix} a_2 & b_2 \\ a_3 & b_3 \end{matrix} \right|}, \, -\frac{\left| \begin{matrix} a_2 & c_2 \\ a_3 & c_3 \end{matrix} \right|}{\left| \begin{matrix} a_2 & b_2 \\ a_3 & b_3 \end{matrix} \right|} \right) \]

\[ B = \left( \frac{\left| \begin{matrix} b_1 & c_1 \\ b_3 & c_3 \end{matrix} \right|}{\left| \begin{matrix} a_1 & b_1 \\ a_3 & b_3 \end{matrix} \right|}, \, -\frac{\left| \begin{matrix} a_1 & c_1 \\ a_3 & c_3 \end{matrix} \right|}{\left| \begin{matrix} a_1 & b_1 \\ a_3 & b_3 \end{matrix} \right|} \right) \]

\[ C = \left( \frac{\left| \begin{matrix} b_1 & c_1 \\ b_2 & c_2 \end{matrix} \right|}{\left| \begin{matrix} a_1 & b_1 \\ a_2 & b_2 \end{matrix} \right|}, \, -\frac{\left| \begin{matrix} a_1 & c_1 \\ a_2 & c_2 \end{matrix} \right|}{\left| \begin{matrix} a_1 & b_1 \\ a_2 & b_2 \end{matrix} \right|} \right) \]

Consider the determinant:

\[ D = \left| \begin{matrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{matrix} \right| \]

Cofactor matrix of \( D \) is:

\[ \text{Cofactor}(D) = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix} \]

Observe that the coordinates \( A, B, \) and \( C \) can be written using the cofactors of the elements in the determinant \( D \) as:

\[ A = \left( \frac{C_{11}}{C_{13}}, \frac{C_{12}}{C_{13}} \right) \]

\[ B = \left( \frac{C_{21}}{C_{23}}, \frac{C_{22}}{C_{23}} \right) \]

\[ C = \left( \frac{C_{31}}{C_{33}}, \frac{C_{32}}{C_{33}} \right) \]

Area of \( \Delta ABC \):

\[ \text{Area} = \frac{1}{2} \left| \begin{matrix} x_A & y_A & 1 \\ x_B & y_B & 1 \\ x_C & y_C & 1 \end{matrix} \right| \]

Substitute the coordinates:

\[ \text{Area} = \frac{1}{2} \left|\left| \begin{matrix} \frac{C_{11}}{C_{13}} & \frac{C_{12}}{C_{13}} & 1 \\ \frac{C_{21}}{C_{23}} & \frac{C_{22}}{C_{23}} & 1 \\ \frac{C_{31}}{C_{33}} & \frac{C_{32}}{C_{33}} & 1 \end{matrix} \right|\right| \]

This can be simplified to:

\[ \text{Area} = \frac{1}{2\left|C_{13}C_{23}C_{33}\right|} \left|\left| \begin{matrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{matrix} \right|\right| \]

We know that:

\[ \left| \text{Cof}(X) \right| = [\text{det}(X)]^2 \]

Therefore:

\[ \text{Area} = \frac{1}{2\left|C_{13}C_{23}C_{33}\right|} \left| \begin{matrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{matrix} \right|^2 \]

Where:

\[ C_{13} = \left| \begin{matrix} a_2 & b_2 \\ a_3 & b_3 \end{matrix} \right| \]

\[ C_{23} = \left| \begin{matrix} a_1 & b_1 \\ a_3 & b_3 \end{matrix} \right| \]

\[ C_{33} = \left| \begin{matrix} a_1 & b_1 \\ a_2 & b_2 \end{matrix} \right| \]

Thus, the area of the triangle formed by the three lines is:

\[ \text{Area} = \frac{1}{2\left|C_{13}C_{23}C_{33}\right|} \left| \begin{matrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{matrix} \right|^2 \]

A nice formula for the area

When the equations of lines are given in the form:

\[ L_1: y = m_1 x + c_1 \]

\[ L_2: y = m_2 x + c_2 \]

\[ L_3: y = m_3 x + c_3 \]

The area of the triangle formed by \( L_1, L_2, \) and \( L_3 \) is:

\[ \frac{1}{2} \left| \sum \frac{(c_1 - c_2)^2}{m_1 - m_2} \right| \]

Area of a parallelogram

Area of a Parallelogram

The area of a parallelogram can be determined when the distances between the pairs of parallel sides are \( d_1 \) and \( d_2 \), and the angle between the two non-parallel sides is \( \theta \). The formula for the area is:

\[ \text{Area} = \frac{d_1 d_2}{|\sin \theta|} \]

Proof:

Consider the parallelogram \(ABCD\). Let \(\angle BAD = \theta\).

The area of the parallelogram is given by the base times the height:

\[ \text{Area of parallelogram} = \text{base} \times \text{height} = AB \times DN \]

Since \( AB = CD \):

\[ \text{Area} = CD \times DN \]

The height \( DN \) can be expressed in terms of the distances \( d_1 \) and \( d_2 \), and the angle \(\theta\):

\[ \text{Area} = \left| \frac{DM \times DN}{\sin \theta} \right| \]

Thus,

\[ \text{Area} = \frac{d_1 \cdot d_2}{|\sin \theta|} \]

Thus, the area of the parallelogram \(ABCD\) is:

\[ \text{Area} = \frac{d_1 d_2}{|\sin \theta|} \]

Area of a Parallelogram Given the Equations of Its Sides

Consider a parallelogram \(ABCD\) with the equations of its sides given by:

• \(AB: ax + by + c_1 = 0\)
• \(BC: dx + ey + f_1 = 0\)
• \(CD: ax + by + c_2 = 0\)
• \(AD: dx + ey + f_2 = 0\)

We aim to find the area of the parallelogram.

Formula for the Area

The area of the parallelogram can be given by:

\[ \text{Area} = \left| \frac{(c_1 - c_2)(f_1 - f_2)}{\left| \begin{matrix} a & b \\ d & e \end{matrix} \right|} \right| \]

Proof:

We know that the area of a parallelogram is given by:

\[ \text{Area} = \left| \frac{d_1 d_2}{\sin \theta} \right| \]

where \(d_1\) and \(d_2\) are the distances between the pairs of parallel sides, and \(\theta\) is the angle between consecutive sides.

1. Calculate \(d_1\) and \(d_2\):

   • The distance between the lines \(ax + by + c_1 = 0\) and \(ax + by + c_2 = 0\) (i.e., \(d_1\)) is:

     \[ d_1 = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \]

   • The distance between the lines \(dx + ey + f_1 = 0\) and \(dx + ey + f_2 = 0\) (i.e., \(d_2\)) is:

     \[ d_2 = \frac{|f_1 - f_2|}{\sqrt{d^2 + e^2}} \]

2. Calculate the Slope of the Sides:

   • The slope of \(AB\) (and \(CD\)) is:

     \[ \text{slope of } AB = -\frac{a}{b} \]

   • The slope of \(AD\) (and \(BC\)) is:

     \[ \text{slope of } AD = -\frac{d}{e} \]

3. Calculate \(\tan \theta\):

   \[ \tan \theta = \frac{\text{slope of } AB - \text{slope of } AD}{1 + (\text{slope of } AB) (\text{slope of } AD)} \]
   \[ \tan \theta = \frac{-\frac{a}{b} - \left(-\frac{d}{e}\right)}{1 + \left(-\frac{a}{b}\right) \left(-\frac{d}{e}\right)} \]
   \[ \tan \theta = \frac{-\frac{a}{b} + \frac{d}{e}}{1 + \frac{ad}{be}} = \frac{bd - ae}{be + ad} \]

4. Calculate \(\sin \theta\):

   \[ \tan \theta = \frac{bd - ae}{be + ad} \Rightarrow \sin \theta = \frac{bd - ae}{\sqrt{(bd - ae)^2 + (be + ad)^2}} \]
   \[ = \frac{bd - ae}{\sqrt{b^2 d^2 + a^2 e^2 - 2 bdae + b^2 e^2 + a^2 d^2 + 2 bdae}} \]
   \[ = \frac{bd - ae}{\sqrt{b^2(d^2 + e^2) + a^2(e^2 + d^2)}} \]
   \[ = \frac{bd - ae}{\sqrt{(d^2 + e^2)(a^2 + b^2)}} \]

5. Substitute \(d_1\), \(d_2\), and \(\sin \theta\) into the Area Formula:

   \[ \text{Area} = \left| \frac{d_1 d_2}{\sin \theta} \right| \]

   Substituting \(d_1\), \(d_2\), and \(\sin \theta\):

   \[ \text{Area} = \left| \frac{\left(\frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}\right) \left(\frac{|f_1 - f_2|}{\sqrt{d^2 + e^2}}\right)}{\left| \frac{bd - ae}{\sqrt{(d^2 + e^2)(a^2 + b^2)}} \right|} \right| \]

   Simplifying:

   \[ \text{Area} = \left| \frac{(c_1 - c_2)(f_1 - f_2)}{(bd - ae)} \right| \]

   Thus, the area of the parallelogram \(ABCD\) is:

   \[ \text{Area} = \left| \frac{(c_1 - c_2)(f_1 - f_2)}{\left| \begin{matrix} a & b \\ d & e \end{matrix} \right|} \right| \]