Check sibling questions

Ex 10.3, 17 - In ABC, vertices A (2, 3), B (4, -1), C (1, 2)

Ex 10.3, 17 - Chapter 10 Class 11 Straight Lines - Part 2
Ex 10.3, 17 - Chapter 10 Class 11 Straight Lines - Part 3 Ex 10.3, 17 - Chapter 10 Class 11 Straight Lines - Part 4 Ex 10.3, 17 - Chapter 10 Class 11 Straight Lines - Part 5 Ex 10.3, 17 - Chapter 10 Class 11 Straight Lines - Part 6 Ex 10.3, 17 - Chapter 10 Class 11 Straight Lines - Part 7

This video is only available for Teachoo black users


Transcript

Ex 10.3, 17 In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A. Let ABC be the triangle with vertex A(2, 3), B(4, –1) & C(1, 2) & AM be the altitude of triangle ABC We need to calculate length & equation of altitude AM Now, Altitude AM is perpendicular to BC We know that if two lines are perpendicular, then the product of their slopes is –1 ∴ Slope of AM × Slope of BC = –1 Slope of AM = (−1)/(𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝐵𝐶) Calculating slope of BC Slope of a line joining points (x1,y1) & (x2,y2) is = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) Slope of BC joining (4, –1) & (1, 2) = (2 − (−1))/(1 − 4) = (2 + 1)/( − 3) = 3/(−3) = –1 From (1) Slope of AM = (−1)/(slope of BC) = (−1)/(−1) = 1 Now, finding equation of altitude AM Altitude AM passes through the point A(2, 3) & has slope 1 The equation of line passing through a point (x1, y1) with slope m is (y – y1) = m (x – x1) Putting values for AM: x1 = 2, y1 = 3, & m = 1 (y – 3) = 1(x – 2) x – y + 1 = 0 1 = y – x y – x = 1 Therefore, equation of the altitude AM is y – x = 1. Also, we have to find length of altitude AM Length of AM = Perpendicular distance from point A to line BC To find perpendicular distance, we need to find equation of line BC Finding equation of line BC Slope of line BC = –1 and it passes through the point B(4,–1) The equation of line passing through a point (x1, y1) with slope m is (y – y1) = m (x – x1) Putting values for BC: x1 = 4, y1 = –1 , m = –1 (y – (-1)) = –1(x – 4) (y + 1) = –1 (x – 4) y + 1 = – x + 4 y + x = 4 – 1 y + x = 3 x + y – 3 = 0 Hence, equation of line BC is x + y – 3 = 0 The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d = |𝐴𝑥_1 + 𝐵𝑦_1 + 𝐶|/√(𝐴^2 + 𝐵^2 ) The above equation is of the form Ax + By + C = 0 Where A = 1 , B = 1 , C = – 3 Also, perpendicular distance from point (2, 3) Hence, x1 = 2, y1 = 3 d = |𝐴𝑥_1 + 𝐵𝑦_1 + 𝐶|/√(𝐴^2 + 𝐵^2 ) Putting values = |1(2) + 1(3) + (−3)|/√((1)2 + (1)2) = |2 + 3 − 3|/√(1 + 1) = |2|/√2 = 2/√2 = 2/√2 × √2/√2 = (2√2)/2 = √2 ∴ Length of altitude AM = √𝟐

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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.