Example 18 - Chapter 10 Class 11 Straight Lines (Term 1)
Last updated at Sept. 6, 2021 by Teachoo
Examples
Example 1 (a)
Example 1 (b)
Example 1 (c) Important
Example 1 (d)
Example 2 Important
Example 3 Important
Example 4
Example 5
Example 6 Important
Example 7
Example 8
Example 9 (i)
Example 9 (ii) Important
Example 10
Example 11 Deleted for CBSE Board 2023 Exams
Example 12
Example 13 Important
Example 14 Important Deleted for CBSE Board 2023 Exams
Example 15 Important
Example 16
Example 17
Example 18 Important You are here
Example 19
Example 20
Example 21 Important
Example 22 Important
Example 23
Example 24 Important
Example 25 Important
Transcript
Example 18 Find the distance of the point (3, ā5) from the line 3x ā 4y ā26 = 0. We know that distance (d) of a point (x1, y1) from a line Ax + By + C = 0 is d = |š“š„_1 + ćšµš¦ć_2 + š¶|/ā(š“^2 + šµ^2 ) Now, our equation is 3x ā 4y ā 26 = 0 The above equation is of the form Ax + By + C = 0 where A = 3, B = ā4 , C = ā26 & we have to find the distance of the point (3, ā 5) from the line So, x1 = 3 , y1 = ā5 Now finding distance d = |š“š„_1 + ćšµš¦ć_2 + š¶|/ā(š“^2 + šµ^2 ) Putting values = |3(3) + (ā4)( ā5) ā 26|/ā(32 + (ā4)2) = |9 + 20 ā 26|/ā(9 + 16) = |29 ā 26|/ā25 = |3|/ā(5 Ć 5) = |3|/5 = 3/5 ā“ Required distance = d = š/š units
