Unformatted text preview: THE POINT GRADIENT FORMULA lfa line passes through a given point (x 1 , y 1) and has a known gradient (mi, :
then the EQUATION of the line can be determined using the formula: y—y1= "105—351) In some textbooks it has not been transposed, and the student might recognise
it in this format: NOTE It is very important to understand and be able to use this formula as its use occurs many times
through the 2 Unit syllabus. Example: Find the equation of a line with gradient 3 passing through the point
(— 2 , 4). Solution: y—4 : 3(x——2)
y—4 : 3x+6
y 2 3x+10. Example: Find the equation ofa line which passes through the point (4, —2)
and is parallel to the line y = 2x+5. If y = 2x+5, then this is ofthe form y = mx+b,
where gradient 2 m = 2, the required line is y — —2 2(x—4)
y+2 — 295—8 2 x — y — 10 O (in general form) ...
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 Winter '10
 golishr

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