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Misc 5 - 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
Misc 5 Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0. First we calculate point of intersection of lines x – 7y + 5 = 0 & 3x + y = 0 Solving (1) x – 7y + 5 = 0 x = 7y − 5 Putting value of x in (2) 3x + y = 0 3(7y − 5) + y = 0 21y − 15 + y = 0 21y + y − 15 = 0 22y − 15 = 0 22y = 15 y = 15/22 Putting value of y = 15/22 in (1) x – 7y + 5 = 0 x – 7(15/22) + 5 = 0 x = 7(15/22) − 5 x = (15 × 7 − 22 × 5)/22 x = (105 − 110)/22 x = ( −5)/22 Hence point of intersection is (( −5)/22, 15/22) The equation of line parallel to y-axis is of the form x = p where p is some constant Given that this equation of line passing through point of intersection (( − 5)/22, 15/22) Hence point (( − 5)/22, 15/22) will satisfy the equation of line Putting x = ( − 5)/22 in the equation x = p ( −5)/22 = p p = ( −5)/22 Thus, Required equation of line is x = p Putting values x = ( −𝟓)/𝟐𝟐
