I'm assuming you mean the point (2,5)
y=mx+b
m=3
x=2
y=5
5=3(2)+b
5=6+b
5/6=b
y=3x+(5/6)
y=mx+b y0=mx0+b 5=3*2+b b=5-5=0 y=3x+0
If you mean a slope of -5 and a point of (6, 3) then the equation is y = -5x+33
If you mean a slope of 23 and a point of (0, 4) then the equation is y = 23x+4
Equation: y-3 = -5(x-6) => y = -5x+33
Equation: y-3 = -5(x-6) => y = -5x+33
y = 2x + 1.
The equation of the line is of the form y = 3x + c where c is a constant. The point (4,9) is on the line, so substituting x=4, y=9 in the equation, 9 = 3*4 + c = 12 + c so c = -3 So the equation of the line is y = 3x - 3
y=mx+b y0=mx0+b 5=3*2+b b=5-5=0 y=3x+0
Choose the equation of the line that contains the points (1, -1) and (2, -2).
If you mean a slope of -5 and a point of (6, 3) then the equation is y = -5x+33
If you mean a slope of 23 and a point of (0, 4) then the equation is y = 23x+4
Write the equation in slope-intercept form of the line that has a slope of 2 and contains the point (1, 1).
Slope: -5 Points: (6, 3) Equation: y = -5x+33
Equation: y-3 = -5(x-6) => y = -5x+33
Equation: y-3 = -5(x-6) => y = -5x+33
Equation: y-3 = -5(x-6) => y = -5x+33
Equation: y-3 = -5(x-6) => y = -5x+33