Last updated at Feb. 3, 2020 by Teachoo
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Example 16 Show that two lines a1x + b1y + c1 = 0 and a2 x + b2 y + c2 = 0 , where b1, b2 โ 0 are: (i) Parallel if ๐_1/๐_1 = ๐2/๐2 The given lines are a1x + b1y + c1 = 0 & a2 x + b2 y + c2 = 0 Let slope of line (1) be m1 & slope of line (2) be m2 If two lines are parallel, then their slopes are equal If line (1) & (2) are parallel , then m1 = m2 Finding m1 & m2 From (1) a1x + b1y + c1 = 0 b1y = โc1 โ a1 x b1y = โa1 x โc1 y = ( โ๐_1 ๐ฅ โ ๐_1)/๐_1 y = ((โ๐_1)/๐_1 ) x โ(๐_1/๐_1 ) The above equation is of the form y = mx + c where m is the slope Thus, Slope of line (1) = m1 = (โ๐_1)/๐_1 From (2) a2x + b2y + c2 = 0 b2y = โc2 โ a2 x b2y = โa2 x โc2 y = ( โ๐_2 ๐ฅ โ ๐_2)/๐_2 y = ((โ๐_2)/๐_2 )x + (๐_2/๐_2 ) The above equation is of the form y = mx + c where m is the slope Thus, Slope of line (2) = m2 = (โ๐_2)/๐_2 Since line (1) & (2) are parallel. So, m1 = m2 (โ๐_1)/๐_1 = (โ๐_2)/๐_2 ( ๐_๐)/๐_๐ = ๐_๐/๐_๐ Hence proved Example 16 Show that two lines a1x + b1y + c1 = 0 and a2 x + b2 y + c2 = 0 , where b1, b2 โ 0 are: (ii) Perpendicular if a1a2 + b1b2 = 0 . If two lines are perpendicular, then product of their slope is equal to โ1 Since line (1) & (2) are perpendicular โ (Slope of line 1) ร (Slope of line 2) = โ1 m1 ร m2 = โ 1 ( โ๐_1)/๐_1 ร ( โ๐_2)/๐_2 = โ1 ( ๐_1)/๐_1 ร ๐_2/๐_2 = โ1 a1a2 = โb1b2 a1a2 + b1b2 = 0 Hence proved
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