Last updated at Jan. 29, 2020 by Teachoo

Transcript

Example 24 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) Point A(α1, β1) lie on line (1) Putting x = α1 & y = β1 in (1) 5α1 − β1 + 4 = 0 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 …(3) 3α1 + 4β1 – 42 = 0 …(4) 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 the 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 – 1 = ((222 − 5(23))/23)/((26 − 23)/23 )(x – 1) y – 1 = (222 − 115)/23× 23/3(x – 1) y – 1 = (107 )/3 (x – 1) 3(y – 1) = 107 (x – 1) 3y – 3 = 107x – 107 107x – 107 = 3y – 3 107x – 3y – 107 + 3 = 0 107 – 3y – 104 = 0 Hence the required equation of line to 107x – 3y – 104 = 0

Examples

Example 1
Important

Example 2

Example 3 Important

Example 4

Example 5

Example 6 Important

Example 7

Example 8

Example 9 Important

Example 10

Example 11

Example 12

Example 13 Important

Example 14 Important

Example 15 Important

Example 16

Example 17

Example 18

Example 19

Example 20

Example 21 Important

Example 22 Important

Example 23

Example 24 Important You are here

Example 25 Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.