Area of triangle bounded by straight line ax + by + c = 0,,,,,(a,b,c ne 0) and coordinate axes is

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9.Straight Line

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

A

$\frac{1}{2}\frac{{{a^2}}}{{|bc|}}$

B

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

C

$\frac{1}{2}\frac{{{b^2}}}{{|ac|}}$

D

$0$

Solution

Given $a x+b y+c=0$

When $x=0, y=-c / b$

When $y=0, x=-c / a$

We got the $x$-intercept as $x=-c / a$

$y$-intercept $y=-c / b$

Area of triangle $=1 / 2 \times$ base $\times$ height

$=1 / 2 \times|(c / a)| \times|(c / b)|$

$=1 / 2\left|c^2 / a b\right|$

Standard 11

Mathematics

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