Points: (20, 18) and (35, 6) Slope: -4/5 Equation: y = -4/5x+34
It is: y = 5x+6
Points: (5, -2) and (4, 3)Slope: -5Equation: y = -5x+23
This starts with the collocation circle to go through the three points on the curve. First write the equation of a circle. Then write three equations that force the collocation circle to go through the three points on the curve. Last, solve the equations for a, b, and r.
You write it as: y = 5x-4 Then you calculate a few sample points, plot them, and draw a straight line through them. Since the equation is linear, two points are enough, in theory, but it is usually recommended to plot a third point, as a verification.
Y=4x+3
Write the equation of the line that passes through the points (3, -5) and (-4, -5)
x = 1
y=mx+c where x and y are variables, m is the gradient (or slope) and c is the intercept on y (axis). that is the general equation of a straight line. if you had given some coordinates for the points one could extrapolate from that to find the full equation. since you have not, one cannot.
Points: (-3, -4) and (6, -1) Slope: 1/3 Equation: 3y = x-9
Points: (20, 18) and (35, 6) Slope: -4/5 Equation: y = -4/5x+34
Slope-intercept form
Points: (2, 5) and (4, 3) Slope: -1 Equation: y = -x+7 in slope-intercept form --- If you want to write the slope-intercept form of the equation of the line passing through the given points, then use the two points to find the slope of the line. After that, use the slope and one of the points to find the y-intercept. For instance, m = (5 - 3)/(2 - 4) = 2/-2 = -1(the slope) y = mx + b (replace m with -1, and (x, y) with (4, 3)) 3 = -1(4) + b 3 = -4 + b (add 4 to both sides) 7 = b Thus, y = -x + 7 is the equation of the line passing through (2, 5) and (4, 3).
It is: y = 5x+6
Points: (5, -2) and (4, 3)Slope: -5Equation: y = -5x+23
the Equation of a Line Given That You Know Two Points it Passes Through.
y=-3x-2