Question

Find the area of the triangle formed by the lines
(i) y = m₁ x + c₁, y = m₂ x + c₂ and x = 0
(ii) y = 0, x = 2 and x + 2y = 3.
(iii) x + y − 6 = 0, x − 3y − 2 = 0 and 5x − 3y + 2 = 0

Answer 1

(i) y = m₁x + c₁ ... (1)

y = m₂x + c₂ ... (2)

x = 0 ... (3)

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2):
$x=\frac{c_2-c_1}{m_1-m_2},y=\frac{m_1{c}_2-m_2{c}_1}{m_1-m_2}$

Thus, AB and BC intersect at B $\left(\frac{c_2-c_1}{m_1-m_2},\frac{m_1{c}_2-m_2{c}_1}{m_1-m_2}\right)$.

Solving (1) and (3):
$x=0,y=c_1$

Thus, AB and CA intersect at A $\left(0,c_1\right)$.

Similarly, solving (2) and (3):
$x=0,y=c_2$

Thus, BC and CA intersect at C $\left(0,c_2\right)$.

∴ Area of triangle ABC = $\frac{1}{2}\left|\begin{array}{ccc}0& c_1& 1\\ 0& c_2& 1\\ \frac{c_2-c_1}{m_1-m_2}& \frac{m_1{c}_2-m_2{c}_1}{m_1-m_2}& 1\end{array}\right|$

= $\frac{1}{2}\left(\frac{c_2-c_1}{m_1-m_2}\right)\left(c_1-c_2\right)=\frac{1}{2}\frac{{\left(c_1-c_2\right)}^2}{m_2-m_1}$

(ii) y = 0 ... (1)

x = 2 ... (2)

x + 2y = 3 ... (3)

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2):
x = 2, y = 0

Thus, AB and BC intersect at B (2, 0).

Solving (1) and (3):
x = 3, y = 0

Thus, AB and CA intersect at A (3, 0).

Similarly, solving (2) and (3):
x = 2, y = $\frac{1}{2}$

Thus, BC and CA intersect at C $\left(2,\frac{1}{2}\right)$.

∴ Area of triangle ABC = $\frac{1}{2}\left|\begin{array}{ccc}2& 0& 1\\ 3& 0& 1\\ 2& \frac{1}{2}& 1\end{array}\right|=\frac{1}{4}$

(iii) x + y − 6 = 0 ... (1)

x − 3y − 2 = 0 ... (2)

5x − 3y + 2 = 0 ... (3)

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2):
x = 5, y = 1

Thus, AB and BC intersect at B (5, 1).

Solving (1) and (3):
x = 2, y = 4

Thus, AB and CA intersect at A (2, 4).

Similarly, solving (2) and (3):
x = −1, y = −1

Thus, BC and CA intersect at C (−1, −1).

∴ Area of triangle ABC = $\frac{1}{2}\left|\begin{array}{ccc}5& 1& 1\\ 2& 4& 1\\ -1& -1& 1\end{array}\right|=12$