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Example 3 - Chapter 9 Class 11 Straight Lines
Last updated at April 16, 2024 by Teachoo
Slope - Prependicularity
Chapter 9 Class 11 Straight Lines
Concept wise
- Cordinate geometry questions
- Slope - Finding slope
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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
- 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
Example 3 Line through the points (–2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x. Let points be A(–2, 6), B(4, 8) , C(8, 12) and D(x, 24) If two lines are perpendicular , then product of their slope is –1 So, Slope of AB × Slope of CD = –1 We know that slope of a line through the points (x1, y1) , (x2, y2)is m = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) Slope of line AB passing through A(– 2, 6) & B(4, 8) Here x1 = −2, y1 = 6 x2 = 4, y2 = 8 Putting avalues Slope of AB = (8 − 6)/(4 − (−2)) = 2/(4 − (−2)) = 2/6 = 1/3 Slope of line CD passing through C(8, 12) & D(x, 24) Here x1 = 8, y1 = 12 x2 = x, y2 = 24 Putting values Slope of CD = (24 − 12)/(𝑥 − 8) = 12/(𝑥 − 8) From (1) Slope of AB × Slope of CD = –1 1/3 × (12/(𝑥 − 8)) = –1 4/((𝑥 − 8)) = –1 4 = –1(x – 8) 4 = –x + 8 x = 8 – 4 x = 4 Thus, value of x is 4
