Example 13 - Chapter 10 Class 11 Straight Lines (Term 1)
Last updated at Feb. 3, 2020 by Teachoo
Transcript
Example 13 Equation of a line is 3x – 4y + 10 = 0. Find its (i) slope 3x – 4y + 10 = 0 −4y = −3x − 10 −4y = − (3x + 10) 4y = 3x + 10 y = (3𝑥 + 10)/4 y = 3/4 x + 10/4 y = 3/4x + 5/2 The above equation is of the form y = mx + c We convert it to y = mx + c where m is the slope & c is the y-intercept Here, m = 3/4 & c = 5/2 Hence, slope m = 𝟑/𝟒 & y-intercept = 5/2 Example 13 Equation of a line is 3x – 4y + 10 = 0. Find its (ii) x and y-intercepts. 3x – 4y + 10 = 0 3x − 4y = − 10 Dividing both sides by − 10 (3𝑥 − 4𝑦)/( −10) = ( −10)/( −10) 3𝑥/( −10) − 4𝑦/( −10) = 1 3𝑥/( −10) + 4𝑦/( 10) = 1 𝑥/(( −10)/3) + 𝑦/(10/4) = 1 𝑥/((−10)/3) + 𝑦/(5/2) = 1 The above equation of the form 𝑥/𝑎 + 𝑦/(𝑏 ) = 1 Hence, a = x-intercept = (−𝟏𝟎)/𝟑 & b = y-intercept = 𝟓/𝟐
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
Example 12
Example 13 Important You are here
Example 14 Important
Example 15 Important
Example 16
Example 17
Example 18 Important
Example 19
Example 20
Example 21 Important
Example 22 Important
Example 23
Example 24 Important
Example 25 Important
Chapter 10 Class 11 Straight Lines (Term 1)
Serial order wise
- Ex 10.1
- Ex 10.2
- Ex 10.3
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Examples
- Miscellaneous
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.