Example 24 - A line is such segment between 5x - y + 4 = 0

Example 24 - Chapter 10 Class 11 Straight Lines - Part 2
Example 24 - Chapter 10 Class 11 Straight Lines - Part 3 Example 24 - Chapter 10 Class 11 Straight Lines - Part 4 Example 24 - Chapter 10 Class 11 Straight Lines - Part 5 Example 24 - Chapter 10 Class 11 Straight Lines - Part 6 Example 24 - Chapter 10 Class 11 Straight Lines - Part 7

This video is only available for Teachoo black users

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


Transcript

Example 15 A line is such that its segment between the lines 5x – y + 4 = 0 and 3x + 4y – 4 = 0 is bisected at the point (1, 5). Obtain its equation. Given lines are 5x – y + 4 = 0 3x + 4y – 4 = 0 Let AB be the segment between the lines (1) & (2) & point P(1, 5) be the mid-point of AB We need to find equation of line AB Let the points be A(α1, β1) & B(α2, β2) Given that Line segment AB is bisected at the point P (1, 5) i.e. P(1, 5) is the mid point of line AB ∴ (1, 5) = (( 𝛼_(1 )+ 𝛼_2 )/2,( 𝛽_(1 )+ 𝛽_2 )/2) a Point A(α1, β1) lie on line (1) Putting x = α1 & y = β1 in (1) 5α1 − β1 + 4 = 0 ( 𝜶_(𝟏 )+ 𝜶_𝟐 )/𝟐 = 1 𝛼1+𝛼2 = 2 𝛼2 = 2 – 𝛼1 ( 𝜷_(𝟏 )+ 𝜷_𝟐 )/𝟐 = 5 𝛽1+𝛽2 = 10 𝛽2 = 10 – 𝛽1 Also, Point B(α2, β2) lie on line (2) Putting x = α2 & y = β2 in (2) 3α2 + 4β2 – 4 = 0 Putting values of α2 , β2 from (A) & (B) 3(2 – α1) + 4(10 – β1) – 4 = 0 6 – 3α1 + 40 – 4β1 – 4 = 0 –3α1 – 4β1 – 4 + 6 + 40 = 0 –3α1 – 4β1 + 42 = 0 0 = 3α1 + 4β1 – 42 3α1 + 4β1 – 42 = 0 Now, our equations are 5α1 − β1 + 4 = 0 3α1 + 4β1 – 42 = 0 From (3) 5α1 − β1 + 4 = 0 5α1 + 4 = β1 β1 = 5α1 + 4 Putting value of β1 in (4) 3α1 + 4β1 – 42 = 0 3α1 + 4(5α1 + 4) – 42 = 0 3α1 + 20α1 + 16 – 42 = 0 23α1 – 26 = 0 23α1 = 26 α1 = 26/23 Putting value of in (3) 5α1 − β1 + 4 = 0 5(26/23) − β1 + 4 = 0 5(26/23) + 4 = β1 𝛽1 = 5(26/23) + 4 𝛽1 = (5 × 26 + 4 × 23)/23 𝛽1 = (130 + 92 )/23 𝛽1 = 222/23 Thus, 𝛼1 = 26/23 & 𝛽1 = 222/23 So, A(𝜶𝟏 , 𝜷𝟏) = A (𝟐𝟔/𝟐𝟑 ", " 𝟐𝟐𝟐/𝟐𝟑) So, our required line AB passes through point A (26/23 "," 222/23) & P (1,5) We know that equation of line joining two point (x1, y1) & (x2, y2) is (y – y1) = (𝑦2 − 𝑦1)/(𝑥2 − 𝑥1) (x – x1) Putting values y – 5 = (222/23 − 5)/(26/23 − 1)(x – 1) y – 5 = ((222 − 5(23))/23)/((26 − 23)/23 ) (x – 1) y – 5 = (222 − 115)/23 × 23/3 (x – 1) y – 5 = (107 )/3 (x – 1) 3(y – 5) = 107 (x – 1) 3y – 15 = 107x – 107 107x – 107 = 3y – 15 107x – 3y – 107 + 15 = 0 107x – 3y – 92 = 0 Hence the required equation of line to 107x – 3y – 92 = 0

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.