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Ex 9.3, 12 - Chapter 9 Class 11 Straight Lines
Last updated at Dec. 16, 2024 by Teachoo
Chapter 9 Class 11 Straight Lines
Concept wise
- Cordinate geometry questions
- Slope - Finding slope
- Slope - Prependicularity
- Angle between two lines by Slope
- Collinearity of 3 points by sliope
- Point Slope form
- Slope-Intercept form
- Intercept form
- Normal form
- Given Statement
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Two lines // or/and prependicular
- Angle between two lines
- Distance - Between two parallel lines
- Distance of a point from a line
- Distance of a point from a line along a line
- Other Type of questions - Area of Triangle formed
- Other Type of questions - 3 lines Concurrent
- Other Type of questions - Image
- Other Type of questions - Mix
Transcript
Ex 9.3, 12 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.
