The area of the triangle bounded by the strai | vedclass

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(B) Given the equation of the line is $ax + by + c = 0$.To find the $x$-intercept,set $y = 0$:$ax + c = 0 \implies x = -\frac{c}{a}$.To find the $y$-i

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$\frac{1}{2} \frac{c^2}{|ab|}$

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(B) Given the equation of the line is $ax + by + c = 0$.To find the $x$-intercept,set $y = 0$:$ax + c = 0 \implies x = -\frac{c}{a}$.To find the $y$-intercept,set $x = 0$:$by + c = 0 \implies y = -\frac{c}{b}$.The vertices of the triangle formed with the coordinate axes are $(0, 0)$,$(-\frac{c}{a}, 0)$,and $(0, -\frac{c}{b})$.The area of the right-angled triangle is given by $\frac{1}{2} 	imes |	ext{base}| 	imes |	ext{height}|$.Area $= \frac{1}{2} 	imes |-\frac{c}{a}| 	imes |-\frac{c}{b}|$.Area $= \frac{1}{2} \left| \frac{c^2}{ab} ight| = \frac{c^2}{2|ab|}$.

The area of the triangle bounded by the straight line $ax + by + c = 0$ $(a, b, c \neq 0)$ and the coordinate axes is

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