In need of help. Ill write it is how it is in the text book.
Find the equation of the line having gradient 3/4, that passes through 7,11.
Express your answer in the form i) ax + by + c =0 and ii) y = mx + c
As one point and the gradient are known, use the formula: y - y1 = m(x-x1)
Plug in numbers is y-11 = 3/4(x-7)
Simplify expressing in the form ax + by + c = 0
y-11 = 3/4(x-7)
4y-44 = 3(x-7)
4y-44=3x -21
Heres where im really baffled next step is
3x-4y+23 = 0
In algebra dont you do what you do to the right what you do to the left???? How does moving a positive 3x in front of the 4y put a minus sign in front of 4y. What am i doing here exactly??
Find the equation of the line having gradient 3/4, that passes through 7,11.
Express your answer in the form i) ax + by + c =0 and ii) y = mx + c
As one point and the gradient are known, use the formula: y - y1 = m(x-x1)
Plug in numbers is y-11 = 3/4(x-7)
Simplify expressing in the form ax + by + c = 0
y-11 = 3/4(x-7)
4y-44 = 3(x-7)
4y-44=3x -21
Heres where im really baffled next step is
3x-4y+23 = 0
In algebra dont you do what you do to the right what you do to the left???? How does moving a positive 3x in front of the 4y put a minus sign in front of 4y. What am i doing here exactly??
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