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Ex 9.1, 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
- 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
Ex 9.1, 5 Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, –4) & B (8, 0). We need to find Slope of a line passing through origin O (0, 0) & mid point of P (0, –4) & B (8, 0) Let M be the mid point of P (0, –4) & B (8, 0) We know that mid point of (x1, y1) and (x2, y2) = ((𝑥1 + 𝑥2 )/2,(𝑦1 + 𝑦2)/2) Mid point M of (0, –4) & (8, 0) = ((0 + 8)/2, ( − 4 + 0)/2) = (8/2, (−4)/2) = (4, –2) We know that slope of line passing through (x1, y1) and (x2, y2) is m = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) Slope of line between points O (0, 0) & M (4, –2) Here, x1 = 0, y1 = 0 x2 = 4, y2 = –2 Putting values m = ( − 2 − 0)/(4 − 0) = ( −2)/4 = ( −1)/2 Hence slope of line is (−𝟏)/𝟐
