y = -3x + 5
y=mx+b y0=mx0+b 5=3*2+b b=5-5=0 y=3x+0
Points: (-5, 9) and (-4, 7) Slope: -2 Equation: y = -2x-1 in slope intercept form
Equation of the straight line: y = -3/5x+2 in slope intercept form
The equations are equivalent.
The equation for the given points is y = x+4 in slope intercept form
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
Points: (8, 10) and (-4, 2) Slope: 2/3 Equation: 3y = 2x+14
Points: (-5, 9) and (-4, 7) Slope: -2 Equation: y = -2x-1 in slope intercept form
If you mean points of: (10, -2) and (20, -12) then it is a straigh line equation in the form of y = -x+8
Equation of the straight line: y = -3/5x+2 in slope intercept form
The equations are equivalent.
Points: (0, 5) and (5, 8)Slope: 3/5Equation: y = 3/5x+5 in slope intercept form
It is: y--3 = 6(x--1) => y = 6x+3 In standard form: 6x-y+3 = 0
The equation for the given points is y = x+4 in slope intercept form
Write the equation in slope-intercept form of the line that has a slope of 2 and contains the point (1, 1).