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11 Cards in this Set
- Front
- Back
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How do you calculate the gradient of a straight line? |
gradient = difference in y / difference in x So a line with points (1,1) and (2,3) would have gradient (3-1)/(2-1)=2 |
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What is the gradient of a straight line: (i) parallel to x (ii) parallel to y |
(i) 0 (ii) undefined |
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What is the relationship between the gradients of two parallel lines? |
They're equal |
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What's the relationship between the gradients of two perpendicular lines? |
m[1]*m[2] = -1 Or the gradient of the perpendicular is equal to the negative of the reciprocal of the original gradient |
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What are the two equations of a straight line? |
y = mx + c y - y[1] = m(x-x[1]) |
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How do you find the distance between two points? |
Using Pythagoras' theorem: AB^2=BC^2+AC^2 So sqrt of squared difference between x coordinates add squared difference between y coordinates AB = √ [ (x[1]–x[2])^2 + (y[1]–y[2])^2 ] |
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How do you find the midpoint of AB? |
Midpoint of AB = ( mean x coordinates, mean of y coordinates ) |
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How would you calculate the equation of a parallel line given a point? |
Calculate gradient then use it and the point you know in y–y[1]=m(x–x[1]) |
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How would you calculate the equation of a perpendicular bisector in form ax+bx+c=0 |
First find the midpoint of original line. Next find the gradient of AB, and from this the perpendicular gradient. Knowing the gradient and a point you can calculate the equation and rearrange accordingly |
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How do you find the point of intersection for two straight lines? |
Using substitution or elimination to find one coordinate, then use this in either equation to find the other. |
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How do you find the coordinates of the intersection between a straight line and a parabola or hyperbola? |
Put both equations in terms of y = , then solve either by factorising or using the quadratic equation. |
