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Find the area of the triangle formed by the straight line 3x – 4y + 12 = 0 with co-ordinate axes. |
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Answer» Given straight line equation is 3x – 4y + 12 = 0 ––––– (1) ⇒ 3x – 4y = –12 ⇒ (3x/-12) - (4y/-12) = 1 ⇒ (x/-4) + (y/3) = 1 X – intercept = – 4 Y – intercept = 3 Area of the triangle formed by the straight line (1) with the co-ordinate axes is (1/2) |(X – intercept) (Y – intercept)| = (1/2) |(–4) (3)| = 6 sq. units. |
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