Example 17 - Line perpendicular to x - 2y + 3 = 0, passing - Examples

  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise
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Example 17 Find the equation of a line perpendicular to the line x − 2y + 3 = 0 and passing through the point (1, – 2). Let equation of line AB be x − 2y + 3 = 0 And let point P be P(1,-2) Let line CD be perpendicular to line AB & passing through point P(1,-2) Lets first calculate slope of line AB x − 2y + 3 = 0 −2y = −x − 3 −2y = − (x + 3) 2y = x + 3 y = (𝑥 + 3)/2 y = x/2 + 3/2 y = 1/2 x + 3/2 The above equation is of the form y = mx + c where m is the slope Thus, slope of line AB = 1/2 We know that product of slope of perpendicular lines is –1 Here, line AB is perpendicular to line CD Slope of line AB × Slope of line CD = –1 1/2 × Slope of line CD = –1 Slope of line CD = –1 × 2/1 Slope of line CD = –2 Thus, line CD has a slope –2 , passes through point P(1, − 2) We know that, equation of line having slope m and passing through point (x1,y1) is y – y1 = m(x – x1) Putting values for line CD, x1 = 1, y1 = –2 & m = –2 y – (–2) = –2 (x – 1) y + 2 = –2(x – 1) y + 2 = –2x + 2 y + 2 + 2x – 2 = 0 y + 2x = 0 y = − 2x Thus, the required equation of line is y = − 2x

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