Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Examples

Example 1 (a)

Example 1 (b)

Example 1 (c) Important

Example 1 (d)

Example 2 Important

Example 3 Important

Example 4 Important

Example 5

Example 6

Example 7 (i)

Example 7 (ii) Important

Example 8

Example 9 Important You are here

Example 10

Example 11

Example 12 Important

Example 13 Important

Example 14

Example 15 Important

Example 16 Important

Question 1 Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 Deleted for CBSE Board 2024 Exams

Question 4 Deleted for CBSE Board 2024 Exams

Question 5 Important Deleted for CBSE Board 2024 Exams

Question 6 Important Deleted for CBSE Board 2024 Exams

Question 7 Important Deleted for CBSE Board 2024 Exams

Question 8 Deleted for CBSE Board 2024 Exams

Question 9 Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

Example 9 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