How to form the equation of the line passing through 2 points, say, A (2,1) and B (5,10) ? First find the gradient of the line. Gradient = (10 − 1)/(5 − 2) = 9/3 = 3. Then the equation of line will be

y − 1 = 3( x − 2) ⇒ y = 3x − 6 + 1 ⇒ y = 3x − 5

or, y − 10 = 3( x− 5) will also leads to y = 3x − 5

Take note that if gradient of a line happens to be zero, then the line is parallel to the x-axis and the equation will be y = c (c being a fixed value). If the gradient of a line happens to be infinity (i.e. line is perpendicular to the x-axis), then the equation of the line will be x = k (k being another fixed value).

*Try this question* :

1) Find the equation of the line that is parallel to y − 2x = 5 and bisects the line joining the points (3,1) and (1, −7).

*Answer :*

y − 2x = 5 ⇒ y = 2x +5, therefore gradient = 2

Mid-pt of (3,1) and (1, −7) is 〈(3 + 1)/2, (1 − 7)/2)〉 = (2, −3)

∴ Equation of line is 〈y − (−3)〉/(x − 2) = 2 ⇒ y + 3 = 2(x − 2) ⇒ y = 2x − 4 − 3 or** y = 2x − 7 **

Notice that this equation must have the ‘2x’ term (as the line is parallel to y = 2x + 5).

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*Related*

Whatever has happened to the very nice equation

(y – y1)/(y2 – y1) = (x – x1)/(x2 – x1) ??

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Dear howardat58,

(y-y1)/(y2-y1)=(x-x1)/(x2-x1) is the same as (y-y1)/(x-x1)=(y2-y1)/(x2-x1). And (y2-y1)/(x1-x1) is the gradient of the line as calculated from the two points (x2,y2) and (x1, y1). The result will be the same.

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