Ex 10.2, 10 - Chapter 10 Class 11 Straight Lines (Term 1)
Last updated at Jan. 11, 2019 by Teachoo
Last updated at Jan. 11, 2019 by Teachoo
Transcript
Ex 10.2, 10 Find the equation of the line passing through ( 3, 5) and perpendicular to the line through the points (2, 5) and ( 3, 6). Let AB be the line passing through (-3, 5) & perpendicular to the line CD through (2, 5) and ( 3, 6) Let Slope of AB = m1 & Slope of CD = m2 Now Line AB is perpendicular to line CD If two lines are perpendicular then product of their slopes are equal to -1 Slope of AB Slope of CD = -1 So, m1 m2 = -1 Slope of line passing through (x, y) & (x2, y2) = ( 2 1)/( 2 1 ) So, Slope of line CD passing through (2, 5) and ( 3, 6) m2 = (6 5)/( 3 2) = ( 1)/( 5) = ( 1)/5 From (1) m1 m2 = -1 m1 (( 1)/5) = -1 m1 = -1 ( 5)/( 1) m1 = 5 Slope of line AB = m1 = 5 Equation of line passing through point (x0, y0) & having slope m (y y0) = m1 (x x0) Equation of line AB passing through (-3, 5)& having slope 5 (y 5) = m1 (x (-3)) (y 5) = 5 (x + 3) y 5 = 5x + 15 5x + 15 y + 5 = 0 5y y + 20 = 0 Hence, the required equation is 5y y + 20 = 0
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