Question

Which is an equation of the line passing through points at (1, 6) and (-2, 1)?

Answer 1

Find the slope of the line, m.
then using the slope and one of the points in point-slope form, which you can simplify to slope-intercept.

Answer 2

*then use the slope and one of the points in point-slope form,

Answer 3

First get the slope.

\[{y_2 - y_1} \over {x_2 - x_1}\]

Then use this formula with any of the coordinate pairs

\[y - y_1 = m(x-x_1)\]

Answer 4

so the answer is what? y = -1/2x +13/2?

Answer 5

y=mx+b

Gradient= m = y1-y2/x1-x2
=
6-1/1+2
4/3

y=4/3x+b

Then replace (1,6) in (x,y)
=
6=4/3*1+b
14/3=b

y=4/3x+14/3

if all calculations are correct

Answer 6

sorry, 5/3
my calculations are wrong

Answer 7

the principal is the same, however...

replace 4 with 5, and redo it, and you should get the answer...

Answer 8

y=5/3x+13/3 should be the answer now

Answer 9

Start with the basic eqn of a line: y=mx+b
You can figure out the slope of the line by calculating\[(y_2-y_1)/(x_2-x_1)\]
Which is 5/3. Then you can plug any of the two points into the line equation to find b:
6=(5/3)(1) + b
Solve for b, which is 13/3
and you get the equation y=5/3 x + 13/3