web analytics

Ex 10.3, 14 - Find coordinates of foot of perpendicular from point (-1, 3) to line 3x-4y-16=0 - Ex 10.3

 

  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise
Ask Download

Transcript

Ex 10.3, 14 Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0. Let the equation of line AB be 3x – 4y – 16 = 0 & Point C be (–1,3) CD is perpendicular to the line AB & we need to find coordinates of point D Let coordinates of point D be (a, b) Also point D(a, b) lies on the line AB i.e. point (a, b) satisfy the equation of line AB 3x – 4y – 16 = 0 Putting x = a & y = b in equation 3a – 4b – 16 = 0 3a – 4b = 16 Also, CD is perpendicular to the line AB And we know that if two lines are perpendicular then product of their slopes are equal to –1 ∴ Slope of CD × Slope of AB = –1 Finding Slope of AB Solving equation of line AB 3x – 4y = 16 3x – 16 = 4y 4y = 3x – 16 y = 3/4x – 16/6 y = 3/4x – 4 The above equation is of the form y = mx + c where m = slope of line ∴ m = 3/4 Slope of line AB = 3/4 Now, Slope of CD × Slope of AB = –1 Slope of CD × 3/4 = –1 Slope of CD = –1 × 4/3 Slope of CD = (−4)/3 Also, Line CD passes through the point C(–1, 3) & D(a, b) So, Slope of CD = (𝑦2 − 𝑦1)/(𝑥2 − 𝑥1) (−4)/3 = (𝑏 − 3)/(𝑎 − ( − 1)) (−4)/3 = (𝑏 − 3)/(𝑎 + 1) –4(a + 1) = 3 (b – 3) –4a – 4 = 3b – 9 –4a – 3b = –9 + 4 –(4a + 3b) = -5 4a + 3b = 5 Now, our equations are 3a – 4b = 16 …(1) 4a + 3b = 5 …(2) From (2) 4a + 3b = 5 4a = 5 – 3b a = (5 − 3𝑏)/4 Putting value of a in (1) 3a – 4b = 16 3((5 − 3𝑏)/4) – 4b = 16 (3(5 − 3𝑏) − 4𝑏(4))/4 = 16 3(5 – 3b) – 16b = 64 15 – 9b – 16 b = 16 × 4 15 – 25b = 64 – 25b = 64 – 15 –25b = 49 b = 49/(−25) b = (−49)/25 Putting value of b = (−49)/25 in (2) 4a + 3b = 5 4a + 3((−49)/25) = 5 4a – 147/25 = 5 4a = 5 + 147/25 4a = (5(25) + 147)/25 4a = (125 + 147)/25 4a = 272/25 a = 272/(25 × 4) a = 68/25 Hence (a, b) = (68/25, (−49)/25) Hence co-ordinates of point D(a, b) =(68/25, (−49)/25)

About the Author

CA Maninder Singh's photo - Expert in Practical Accounts, Taxation and Efiling
CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
Jail