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The General Equation of a straight line

The General Equation of a straight line. There are 2 main ways of writing the equation of a straight line:. The gradient/intercept form y = mx + c m = gradient c = y -intercept. General form ax + by + c = 0 a > 0 a, b, and c are * whole numbers

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The General Equation of a straight line

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  1. The General Equation of a straight line There are 2 main ways of writing the equation of a straight line: The gradient/intercept form y = mx + c m = gradient c = y-intercept General form ax + by + c = 0 a > 0 a,b, and c are * whole numbers * “reduced” to lowest integers

  2. Example 1 Write y = 3x + ½ in general form. y = 3x + ½ 2y = 6x + 1 6x– 2y + 1 = 0 Example 2 What is the gradient and y-intercept of the line 8x+ 4y – 12 = 0? Is the line 8x+ 4y – 12 = 0 in general form? 8x+ 4y – 12 = 0 4y = –8x + 12 y = –2x + 3 m = –2 and c = 3

  3. y x (y2y1) (x2, y2) (x2x1) (x1, y1) Gradient between two points The gradient of a line is a measurement of its steepness. The gradient is the ratio of the rise to the run of the line. We use the letter m for gradient. For any line. Pick 2 points (x1, y1) and(x2, y2). Draw a right angled triangle Then use “change in y” over “change in x”

  4. Gradient between two points The gradient of a line may be small, having a “gentle” slope The gradient of a line may be large, having a “steep” slope A line with a positive gradient slopes “upwards” from left to right. A line with a negative gradient slopes “downwards” from left to right.

  5. Example 2 Show the points A(2, 5), B(6, 11) and C(–6, –7) lie on a straight line Mathematical word for this is “collinear” First calculate the gradient of any 2 points, say AB. Then calculate the gradient of either AC or BC What would happen if we had the points the other way around?. As mAB = mAC A, B and C are collinear.

  6. The point gradient formula The equation of a straight line through a given point with a given gradient can be found using: y– y1= m(x – x1) m = gradient (x1, y1) is a point on the line.

  7. Example 1 What is the equation of the line through the point (2, 7) with a gradient of ¾? y – y1 = m(x – x1) y – 7 = ¾(x – 2) 4y – 28 = 3(x – 2) 4y – 28 = 3x – 6

  8. Example 2 What is the equation of the line through the point (3, –4) with a gradient of –5? y – y1 = m(x – x1) y – (–4) = –5(x – 3) y + 4 = –5x + 15 y = –5x + 11

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