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Example 13 - Chapter 10 Class 11 Straight Lines (Term 1)
Last updated at March 22, 2023 by Teachoo
Examples
Example 1 (a)
Example 1 (b)
Example 1 (c) Important
Example 1 (d)
Example 2 Important
Example 3 Important
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Example 6 Important
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Example 9 (i)
Example 9 (ii) Important
Example 10
Example 11 Deleted for CBSE Board 2023 Exams
Example 12
Example 13 Important You are here
Example 14 Important Deleted for CBSE Board 2023 Exams
Example 15 Important
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Example 17
Example 18 Important
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Example 21 Important
Example 22 Important
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Example 25 Important
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
