Last updated at May 29, 2018 by Teachoo

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Ex10.2, 11 A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1 : n. Find the equation of the line. Let line CD perpendicular to the line segment AB joining two points A(1, 0) and B(2, 3) i.e. CD AB We want to equation of line CD We know that if two lines are perpendicular then product of their slope is equal to 1 Hence, Slope of AB Slope of CD = 1 Finding Slope of AB Slope of line joining two points (x1, y1) & (x2, y2) = _(2 _2 )/( _2 _1 ) Slope of line joining two point A(1, 0) and B(2, 3) is Slope of AB = (3 0)/(2 1) = 3/1 = 3 Now, Slope of AB Slope CD = 1 3 Slope of CD = 1 Slope of CD ( 1)/3 Also given that line CD divides AB in the ratio of 1 : n i.e. P divides AB in the ratio 1 : n We know that, If a point divide any line joining (x1, y1) & (x2, y2) in the ratio of m1 : m2 then co-ordinate of that point is (( _2 _2 + _1 _1)/( _1 + _2 ),( _2 _(2 ) + _1 _1)/( _1 + _2 )) P divide AB joining (1, 0) & (2, 3) divide in the ratio of 1 : n Coordinate of P are = ((1 2 + 1 )/(1 + ), (1 3 + 0 )/(1 + )) = ((1 2 + 1 )/(1 + ), (1 3 + 0 )/(1 + )) = ((2 + )/(1 + ), 3/(1 + )) = (( + 2)/(1 + ), 3/(1 + )) Hence P = (( + 2)/(1 + ), 3/(1 + )) Now, we need to find equation of line CD Equation of line passing through point (x0, y0) & having slope m is (y y0) = m(x x0) Equation of line CD passing through point P(( + 2)/( + 1) ", " 3/( + 1)) & having slope ( 1)/3 is ( 3/( + 1)) = ( 1)/3 ( ( + 2)/( + 1)) ( ( + 1) 3)/( + 1) = ( 1)/3 (( ( + 1) ( + 2))/( + 1)) (y(n + 1) 3)/(n + 1) = ( 1)/3 ((x(n + 1) (n + 2))/(n + 1)) 3(y(n + 1) 3) = (x(n + 1) (n + 2)) ((n + 1)/(n + 1)) 3(y(n + 1) 3) = (x(n + 1) (n + 2)) 3y (n + 1) 9 = x (n + 1) + (n + 2) 3y (n + 1) + x(n + 1) = n + 2 + 9 3y (n + 1) + x(n + 1) = n + 11 x(n + 1) + 3(n + 1)y = n + 11 Which is the required equation

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.