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Ex 10.3, 13 - Find equation of right bisector of line joining - Ex 10.3

  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise
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Ex 10.3, 13 Find the equation of the right bisector of the line segment joining the points (3, 4) and (โ€“1, 2). Let AB be the line joining points A(โˆ’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 ((๐‘ฅ1 + ๐‘ฅ2)/2, (๐‘ฆ1 + ๐‘ฆ2)/2) So, co-ordinates of point P = (( โˆ’ 1 + 3)/2, (2 + 4)/2) = (2/2 " , " 6/2) = (1, 3) Since CD is the right bisector of line AB โˆด Line CD โŠฅ line AB And, we know that if two lines are perpendicular, their product of slope is โ€“1 So, Slope of CD ร— Slope of AB = โˆ’ 1 Slope of CD = ( โˆ’ 1)/(๐‘†๐‘™๐‘œ๐‘๐‘’ ๐‘œ๐‘“ ๐ด๐ต) Finding slope of line AB Slope of a line joining points (x1,y1) & (x2,y2) is = (๐‘ฆ_2 โˆ’ ๐‘ฆ_1)/(๐‘ฅ_2 โˆ’ ๐‘ฅ_1 ) Slope of AB joining (3, 4)& ( โˆ’ 1, 2) = (2 โˆ’ 4)/( โˆ’ 1 โˆ’ 3) = (2 โˆ’ 4)/( โˆ’ 1 โˆ’ 3) = ( โˆ’ 2)/( โˆ’ 4) = 1/2 So, Slope of AB = 1/2 โˆด Slope of CD = ( โˆ’ 1)/(๐‘†๐‘™๐‘œ๐‘๐‘’ ๐‘œ๐‘“ ๐ด๐ต) = ( โˆ’ 1)/(1/2) = โˆ’ 2 We know that equation of a line passes through (x1, y1) & having slope of m is (y โˆ’ y1) = m(x โˆ’ x1) Equation of line CD passing through point P(1, 3) & slope of โ€“2 is (y โ€“ 3) = โ€“2 (x โ€“ 1) y โ€“ 3 = โ€“2x + 2 y + 2x = 2 + 3 2x + y = 5 Thus, 2x + y โˆ’ 5 = 0 is the required equation.

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