Use point-slope form to write the equation of a line that has a slope of 2/3 and passes through (-3,-1). Write your final equation in slope-intercept form.
your equation must take the form
y = (2/3)x + b
plug in your given point values to find b
let me know what you get
or, since you are given a point and a slope (and told to use the point-slope form), start with the point-slope form of the equation:
y+1 = 2/3 (x+3)
now rearrange stuff to the form you need.
To write the equation of a line using the point-slope form, we need the slope of the line and a point through which the line passes. In this case, the slope is 2/3 and the line passes through the point (-3, -1).
The point-slope form of a linear equation is given as:
(y - y1) = m(x - x1)
Where (x1, y1) represents the coordinates of the given point, and 'm' represents the slope.
Substituting the given values, we have:
(y - (-1)) = (2/3)(x - (-3))
Simplifying that equation, we get:
(y + 1) = (2/3)(x + 3)
To convert this equation to slope-intercept form (y = mx + b), where 'b' is the y-intercept, we need to isolate y on one side of the equation.
Distributing the (2/3) to (x + 3), we get:
y + 1 = (2/3)x + 2
Subtracting 1 from both sides of the equation to isolate 'y', we obtain:
y = (2/3)x + 2 - 1
Simplifying further:
y = (2/3)x + 1
Therefore, the equation of the line in slope-intercept form is y = (2/3)x + 1.