       1. Chapter 10 Class 11 Straight Lines (Term 1)
2. Serial order wise
3. Ex 10.3

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 = 3/4 x – 16/6 y = 3/4 x – 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 = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) (−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 4a + 3b = 5 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) 