Find the equation of a straight line whuch passes through the points (2,3) and (-1,4)
slope = (4-3)/(-1-2) = -1/3
using that slope and the point (2,3)
y-3 = (-1/3)(x-2)
arrange into whatever version of line you learned.
Sub (-1,4) into your equation to make sure it works
y=mx+c......
m=y2-y1/x2-x1
m=4-3/-1-2
m=-1/3
at any point i pick
p(2,3)
y=mx+c
3=(-1)2/3+c
-9=-2/3+c
-9+2=3c
-9+2/3=c
c=-7/3
now
y=mx+c.......crack out the equation
not quite. You got m right, but
3 = (-1/3)(2)+c
3 = -2/3 + c
3 + 2/3 = c
11/3 = c
y = -1/3 x + 11/3
why did you change 9 to -9? It didn't change sides.
If you check Reiny's equation
y-3 = (-1/3)(x-2)
y-3 = -1/3 x + 2/3
y = -1/3 x + 11/3
You may want to work on logs and advanced algebra, but maybe you should work on the basic skills some more for now...
i just did i without solving ob paper it was a typo okay
To find the equation of a straight line that passes through two points, we can use the slope-intercept form of a linear equation, which is:
y = mx + b
where m is the slope of the line, and b is the y-intercept.
Step 1: Find the slope (m)
The slope (m) can be calculated using the formula:
m = (y2 - y1) / (x2 - x1)
Given the points (2,3) and (-1,4), we can substitute their coordinates into the formula:
m = (4 - 3) / (-1 - 2)
m = 1 / -3
m = -1/3
Step 2: Find the y-intercept (b)
To find the y-intercept (b), we can choose any of the two points and substitute the values into the equation y = mx + b. Let's use the point (2,3):
3 = (-1/3) * 2 + b
3 = -2/3 + b
To isolate b, we can add 2/3 to both sides:
3 + 2/3 = b
9/3 + 2/3 = b
11/3 = b
Step 3: Write the equation
Now that we have the slope (m = -1/3) and the y-intercept (b = 11/3), we can write the equation of the straight line:
y = (-1/3)x + 11/3
So the equation of the straight line that passes through the points (2,3) and (-1,4) is y = (-1/3)x + 11/3.