find the equation of a line with a slope of -2/3 which passes through the point 3,-1?
The equation of a straight line is: y = mx + c
where m is the slope and c is the y-intercept.
Since m = -2/3, our equation then is: y = -(2/3)x + c
We know thast the point (3, -1) satisfies the equation of the straight line.
Therefore, we can substitute for x = 3, and y = -1, into that equation. This gives us,
-1 = -(2/3)*3 + c
-1 = -2 + c
1 = c
The equation of the straight line then is: y = -(2/3)x + 1