Find the equation of a line which passes through the point (3, -1) with the gradient M=2÷3
1 answer:
Step-by-step explanation:
Equ of line -
y=mx+c
Using point (3,-1)
3 as x, - 1 as y
M=2/3
Substitute the values of these coordinates
3=2/3(-1) +c - - - - now find for c
c= 3- (-2/3)
c= 3.67
Now with the value of c, put the values back into the equation form .. Assuming we don't have a point to work with
y=2/3x + 3.67
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◆ Lines n Angles ◆
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