Q.Show that the area of the triangle formed by the lines $y = m_1 x + c_1$, $y = m_2 x + c_2$ and $x = 0$ is $\dfrac{(c_1 - c_2)^2}{2\,|m_1 - m_2|}$.

Area of a Triangle from Lines – First Principles

Imagine you're given three straight lines on a plane. They aren't parallel to each other, so they intersect in three distinct points. Those three intersection points form a triangle. The question is: can you find the area of that triangle directly from the equations of the lines, without first finding the coordinates of the vertices?

That's exactly what "area of triangle from lines" is about. It's a shortcut that saves you from solving three pairs of equations and then plugging into the area formula.

The Intuition

Every line equation can be written in the form $ax + by + c = 0$. If you have three such lines:

The three intersection points are where each pair of lines meets. The area of the triangle formed by these three points can be expressed directly in terms of the coefficients $a_i, b_i, c_i$ — no vertex coordinates needed.

Why does this work? Because the determinant that gives the area of a triangle from its vertices can be rewritten, using the line equations, into a single determinant involving only the coefficients. It's a neat algebraic trick that relies on the fact that each vertex satisfies two of the three line equations.

The Precise Statement

Where each $2\times2$ determinant in the denominator is:

This is the area of the triangle formed by the three lines, assuming no two are parallel (so none of the denominator determinants is zero).

How to Use It – Step by Step

1. Write each line in the form $ax + by + c = 0$. Make sure all three are in the same format — if a line is given as $y = mx + d$, rewrite it as $mx - y + d = 0$ (or equivalently $mx - y + d = 0$).

2. Form the $3\times3$ determinant of all coefficients $a_i, b_i, c_i$ and compute its value. Call it $D$.

3. Compute the three $2\times2$ determinants for each pair of lines:

   • $D_{12} = a_1 b_2 - a_2 b_1$
   • $D_{23} = a_2 b_3 - a_3 b_2$
   • $D_{31} = a_3 b_1 - a_1 b_3$

4. Plug into the formula:

The absolute value in the denominator ensures the area is positive. The numerator is squared, so it's always non-negative.

Why This Formula Works (Briefly)

The standard area formula for a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is:

Now, each vertex lies on two lines. For example, vertex $P_{12}$ (intersection of $L_1$ and $L_2$) satisfies $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$. Using Cramer's rule, you can express $x$ and $y$ of that vertex in terms of the coefficients. Substituting these into the vertex determinant and simplifying yields the formula above. The squared numerator and product of $2\times2$ determinants emerge naturally from the algebra.

Example

Find the area of the triangle formed by the lines:

• $L_1: 2x + 3y - 6 = 0$
• $L_2: x - y + 1 = 0$
• $L_3: 3x + 2y - 12 = 0$

Step 1: Coefficients:

• $a_1=2, b_1=3, c_1=-6$
• $a_2=1, b_2=-1, c_2=1$
• $a_3=3, b_3=2, c_3=-12$

Step 2: $3\times3$ determinant:

Compute:

$= 2[(-1)(-12) - (1)(2)] - 3[(1)(-12) - (1)(3)] + (-6)[(1)(2) - (-1)(3)]$

$= 2[12 - 2] - 3[-12 - 3] -6[2 + 3]$

$= 2(10) - 3(-15) -6(5) = 20 + 45 - 30 = 35$

Step 3: $2\times2$ determinants:

• $D_{12} = (2)(-1) - (3)(1) = -2 - 3 = -5$
• $D_{23} = (1)(2) - (-1)(3) = 2 + 3 = 5$
• $D_{31} = (3)(3) - (2)(2) = 9 - 4 = 5$

Step 4: Area: …