Last updated at May 29, 2018 by Teachoo

Transcript

Ex10.2, 15 The perpendicular from the origin to a line meets it at the point ( 2, 9), find the equation of the line. Let line OC be perpendicular to line AB at point C( 2, 9) i.e. OC perpendicular AB We know that If two lines are perpendicular then product of their slopes is equal to -1 Slope of OC Slope of AB = -1 Finding slope of OC Slope of line passing through (x1, y1) & (x2, y2) = ( 2 1 )/( 2 1) Slope of OC passing through (0, 0) and ( 2, 9) Slope of OC = (9 0)/( 2 0) = 9/( 2) = ( 9)/2 Now, Slope of AB Slope of OC = -1 Slope of AB ( 9/2) = -1 Slope of AB = 2/9 Now, Point C(-2, 9) lie on the line AB Hence line AB passes through C(-2, 9) with slope 2/9 We know that equation of line passing through (x0, y0) & having slope m is (y y0) = m (x x0) Equation of line AB passing through (-2, 9)& having slope 2/9 (y 9) = 2/9 (x (-2)) y 9 = 2/9(x + 2) 9(y 9) = 2(x + 2) 9y 81 = 2x + 4 9y 2x 81 4 = 0 9y 2x 85 = 0 0 = 2x 9y + 85 2x 9y + 85 = 0 Which is the required equation of AB

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.