Find the area of the triangle formed by the lines

$(i) y = m_{1} x + c_{1}, y = m_{2} x + c_{2} a n d x = 0$ $(i i) y = 0, x = 2 a n d x + 2 y = 3.$ $(i i i) x + y - 6 = 0, x - 3 y - 2 = 0 a n d 5 x - 3 y + 2 = 0$

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Solution

$(i) y = m_{1} x + c_{1} . . . (1)$

$y = m_{2} x + c_{2} . . . (2)$

$x = 0 . . . (3)$

$Solving 1 and 2 gives x = (\frac{c_{2} - c_{1}}{m_{1} - m_{2}}),$

$y = (\frac{m_{1} c_{2} - m_{2} c_{1}}{m_{1} - m_{2}})$

$Similarly,$

$Solving 2 and 3 gives (0, c_{2})$

$Thus BC and CA intersect at C (0, c_{2})$

$Solving 1 and 3 gives (0, c_{1})$

$Thus AB and CA intersect at A (0, c_{1})$

$Area of triangle formed by above vertices is$

$= \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & c_{1} & 1 0 & c_{2} & 1 \frac{c_{2} - c_{1}}{m_{1} - m_{2}} & \frac{m_{1} c_{2} - m_{2} c_{1}}{m_{1} m_{2}} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= \frac{1}{2} [(\frac{c_{2} - c_{1}}{m_{1} - m_{2}} \times c_{1}) - (\frac{c_{2} - c_{1}}{m_{1} - m_{2}} \times c_{2})]$

$= \frac{(c_{2} - c_{1})^{2}}{2 (m_{1} - m_{2})}$

$(i i) y = 0, x = 2 a n d x + 2 y = 3$

$y = 0 . . . (1)$

$x = 2 . . . (2)$

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

$Solving (1) and (2)$

$(x, y) = (2, 0) . . . (A)$

$solving (2) and (3)$

$2 + 2 y = 3$

$\Rightarrow y = \frac{1}{2}$

$\Rightarrow x = 2$

$∴ (2, \frac{1}{2}), = (x, y) . . . (B)$

$Solving (1) and (3)$

$x + 0 = 3$

$\Rightarrow Point is (3, 0) . . . (C)$

$Area of triangle is$

$= \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 & 0 & 1 3 & 0 & 1 2 & \frac{1}{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$and treating the points A, B, C as$

$(x_{1} - y_{1}), (x_{2} - y_{2}) a n d (x_{3} - y_{3})$

$= \frac{1}{2} [2 (\frac{1}{2} - 0) + 2 (0 - 0) + 3 (0 - \frac{1}{2})]$

$= \frac{1}{2} [1 - \frac{3}{2}]$

$= \frac{- 1}{4}$

$(i i i) x + y - 6 = 0, x - 3 y - 2 = 0 a n d 5 x$

$- 3 y + 2 = 0$

$x + y - 6 = 0 . . . (1)$

$x - 3 y - 2 = 0 . . . (2)$

$5 x - 3 y + 2 = 0 . . . (3)$

$Solving 1 and 2 gives us (x_{1}, y_{1}) = (5, 1)$

$Solving 2 and 3 gives us (x_{2}, y_{2}) = (- 1 - 1)$

$Solving 1 and 3 gives us (x_{3}, y_{3}) = (2, 4)$

$So Area of triangle when three vertices are given is$

$= \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 5 & 1 & 1 2 & 4 & 1 - 1 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣$

$\frac{1}{2} (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}))$

$= \frac{1}{2} [| - 25 - 3 + 4 |]$

$= 12 sq. units$

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Similar questions

Find the area of the triangle formed by the lines $y = m_{1} x + c_{1}, y = m_{2} x + c_{2}$ and x = 0

x = 0, y = m_{1} x + c_{1}

y = m_{2} x + c_{2}

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