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Question 3 - Collinearity of 3 points by sliope - Chapter 9 Class 11 Straight Lines
Last updated at April 16, 2024 by Teachoo
Collinearity of 3 points by sliope
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
- 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
Question 3 Consider the given population and year graph. Find the slope of the line AB and using it, find what will be the population in the year 2010? The two points given are A (1985, 92), B (1995, 97) Let plot a third point on line C (2010, y) Points A,B, C lie on the line So, A, B & C are collinear ∴ Slope of AB = Slope of BC 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 joining points A(1985, 92) & B(1995, 97) m = (97 − 92)/(1995 − 1985) = 5/10 = 1/2 Slope of BC joining points B(1995, 97) & C(2010, y) m = (𝑦 − 97)/(2010 − 1995) = (𝑦 − 97)/15 Thus, Slope of Line AB is 𝟏/𝟐 Now, Slope of AB = Slope of BC 1/2 = (𝑦 − 97)/15 15/2 = y – 97 7.5 = y – 97 7.5 + 97 = y y = 104.5 Hence, population in the year 2010 will be 104.5 crores
