write an equation in standard form of the line containing (-1,1) and (6,-1)
(-1,1)
(6, -1)
---------
(7,-2) -----> m= -2/7
and then u plug it in the point slope form?
somehow i feel im doing something wrong...
0.00000024
What is the standard form x^2-13x+20=-6
To write the equation of a line in standard form, you need to know the slope (m) and the y-intercept (b) of the line. In this case, you correctly found the slope of the line through the points (-1,1) and (6,-1) as m = -2/7.
Now, let's find the y-intercept (b). We can use the point-slope form of the equation of a line: y - y1 = m(x - x1). Plugging in the values (-1,1) and the slope (-2/7), we have:
y - 1 = (-2/7)(x - (-1))
y - 1 = (-2/7)(x + 1)
y - 1 = (-2/7)x - 2/7
Now, let's simplify the equation:
y - 1 = (-2/7)x - 2/7
y = (-2/7)x - 2/7 + 1
y = (-2/7)x - 2/7 + 7/7
y = (-2/7)x + 5/7
So, the equation of the line in point-slope form is y = (-2/7)x + 5/7.
To convert this equation into standard form, we want the equation to be in the form Ax + By = C, where A, B, and C are integers and A is positive.
Multiply the entire equation by 7 to eliminate the fractions:
7y = -2x + 5
Rearrange the equation to place x and y terms on one side and the constant term on the other side:
2x + 7y = 5
Since the coefficients A and B are both integers, and A is positive, this equation is in standard form. Therefore, the equation in standard form for the line containing the points (-1,1) and (6,-1) is 2x + 7y = 5.