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Ex 10.2, 10 - Line passing through (-3, 5) and perpendicular - Two lines // or/and prependicular

  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise
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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) = (𝑦2 − 𝑦1)/(𝑥2 − 𝑥1 ) 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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