Example 18 - Chapter 10 Class 11 Straight Lines

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Example 18 Find the distance of the point (3, 5) from the line 3x 4y 26 = 0. We know that distance (d) of a point (x1, y1) from a line Ax + By + C = 0 is d = | _1 + _2 + |/ ( ^2 + ^2 ) Now, our equation is 3x 4y 26 = 0 The above equation is of the form Ax + By + C = 0 where A = 3 , B = 4 , C = 26 & we have to find the distance of the point (3, 5) from the line So, x1 = 3 , y1 = -5 Now finding distance d = | _1 + _2 + |/ ( ^2 + ^2 ) Putting values = |3(3) + ( 4)( 5) 26|/ (32 + ( 4)2) = |9 + 20 26|/ (9 + 16) = |29 26|/ 25 = |3|/ (5 5) = |3|/5 = 3/5 Required distance = d = 3/5 units