Write the equation of the line in stasndard form that passes through the points 2,-3 and -1,-4
slope = m = (-4+3)/(-1-2)
= -1/-3 = 1/3
-3 = (1/3)(2) + b
b = -9/3 - 2/3 = -11/3
y = (1/3) x -11/3
3 y = x - 11
x -3 y = 11
I don't understand how you worked to get the y intercept.
I knew the slope m
so I used y = m x + b
I put in 2 for x and -3 for y
that is the point (2,-3)
-3 = (1/3)(2) + b
To write the equation of a line in standard form, we need to find the slope and the y-intercept.
The formula to find the slope (m) is:
m = (y2 - y1) / (x2 - x1)
Let's calculate the slope using the given points (2, -3) and (-1, -4):
m = (-4 - (-3)) / (-1 - 2)
m = (-4 + 3) / (-1 - 2)
m = -1 / -3
m = 1/3
So the slope of the line is 1/3.
Now that we have the slope, we can find the y-intercept (b) using the slope-intercept form of a line (y = mx + b) and one of the given points. Let's use the point (2, -3):
-3 = (1/3) * 2 + b
-3 = 2/3 + b
b = -3 - 2/3
b = -9/3 - 2/3
b = -11/3
So the y-intercept (b) is -11/3.
Now we can write the equation of the line in slope-intercept form (y = mx + b):
y = (1/3)x - 11/3
To convert the equation to standard form, we need to get rid of the fractions. Multiply the entire equation by 3 to eliminate the fractions:
3y = x - 11
Now, rearrange the equation to have the x and y terms on the same side and the constant on the other side:
x - 3y = 11
Therefore, the equation of the line in standard form that passes through the points (2, -3) and (-1, -4) is x -3y = 11.