Ex 10.3
Ex 10.3, 1 (ii) Important
Ex 10.3, 1 (iii)
Ex 10.3, 2 (i)
Ex 10.3, 2 (ii)
Ex 10.3, 2 (iii) Important
Ex 10.3, 3 (i)
Ex 10.3, 3 (ii)
Ex 10.3, 3 (iii) Important
Ex 10.3, 4
Ex 10.3, 5 Important
Ex 10.3, 6 (i) Important
Ex 10.3, 6 (ii)
Ex 10.3, 7
Ex 10.3, 8 Important
Ex 10.3, 9 Important
Ex 10.3, 10
Ex 10.3, 11
Ex 10.3, 12 Important
Ex 10.3, 13 You are here
Ex 10.3, 14 Important
Ex 10.3, 15
Ex 10.3, 16 Important
Ex 10.3, 17 Important
Ex 10.3, 18 Important
Ex 10.3
Last updated at May 29, 2018 by Teachoo
Ex 10.3, 13 Find the equation of the right bisector of the line segment joining the points (3, 4) and ( 1, 2). Let AB be the line joining points A( 1, 2) & B(3, 4) Let CD be the right bisector of line AB We have to find equation of line CD Since CD is the right bisector of line AB, Point P is the mid-point of line AB We know that co-ordinates of mid-point is given by (( 1 + 2)/2, ( 1 + 2)/2) So, co-ordinates of point P = (( 1 + 3)/2, (2 + 4)/2) = (2/2 " , " 6/2) = (1, 3) Since CD is the right bisector of line AB Line CD line AB And, we know that if two lines are perpendicular, their product of slope is 1 So, Slope of CD Slope of AB = 1 Slope of CD = ( 1)/( ) Finding slope of line AB Slope of a line joining points (x1,y1) & (x2,y2) is = ( _2 _1)/( _2 _1 ) Slope of AB joining (3, 4)& ( 1, 2) = (2 4)/( 1 3) = (2 4)/( 1 3) = ( 2)/( 4) = 1/2 So, Slope of AB = 1/2 Slope of CD = ( 1)/( ) = ( 1)/(1/2) = 2 We know that equation of a line passes through (x1, y1) & having slope of m is (y y1) = m(x x1) Equation of line CD passing through point P(1, 3) & slope of 2 is (y 3) = 2 (x 1) y 3 = 2x + 2 y + 2x = 2 + 3 2x + y = 5 Thus, 2x + y 5 = 0 is the required equation.