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Ex 9.3, 6 - Chapter 9 Class 11 Straight Lines
Last updated at April 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
Ex10.3, 7 Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (–2, 3). Let the equation of line AB be 3x – 4y + 2 = 0 & let point P be P(–2, 3) Let line CD be parallel to line AB & passing through the point P(–2,3) We have to find the equation of line CD We know that if two lines are parallel then their slopes are equal , Since line CD is parallel to line AB Thus, Slope of line CD = Slope of line AB Finding slope of line AB 3x − 4y + 2 = 0 3x + 2 = 4y (3𝑥 + 2)/4 = y (3/4)x + 2/4 = y (3/4)x + 1/2 = y y = (3/4)x + 2/4 This equation is of the form y = mx + c where m is the slope, Thus, Slope of line AB = 3/4 Now, from (1) Slope of line CD = Slope of line AB ∴ Slope of line CD = 3/4 Equation of a line passing through a point (x0, y0) & having slope m is (y – y0)= m(x – x0) Equation of a line CD passing through point P(-2,3) & having slope 3/4 is (y – y0)= m(x – x0) Putting values (y − 3) = 3/4(x – (–2)) (y − 3) = 3/4(x + 2) 4(y − 3) = 3 (x + 2) 4y − 12 = 3x + 6 − 12 − 6 = 3x − 4y 3x − 4y = − 12 − 6 3x − 4y = − 18 3x + 4y + 18 = 0 Thus, the required equation of line is 3x + 4y + 18 = 0
