The points of a line are (1,7) (0,3) (-1,-1) and (-2,-5) how do I find it's equation?

Answer 1

m = (3 - 7) / (0 - 1)
m = (- 4) / (-1)
m = 4
Passes thro` (1,7)

y - 7 = 4 (x - 1)
y = 4x - 4 + 7
y = 4x + 3 is required equation.

Answer 2

A straight line ca be described as all points satisfying the equation

y=ax+b

where a is the rate of clibming and b is the y-component of the point where the line and the y-axis cross.


Start by finding the rate of climbing:

a=(y_2 - y_1) / (x_2 - x_1) = 4/1 = 4

find the y-component of the point where the line crosses the y-axis. This happens for x=0. The point (0,3) is that point --> b=3


y= ax+b = 4x+3

Observe that you only need two points to define a line. You have 4. Any selection of these 4 points will give you the same a and b.

Answer 3

If it is a line, then the slope should be the same between any of these pairs of points:

(7 - 3)/(1 - 0) = 4
(3 - (-5)/(0 - (-2) = 4

Ok. So, the "slope/intercept" form of the equation of the line is:

y = mx + b, where "m" is the slope, and "b" is the y-intercept.

Since we already know the slope (4), the equation is:

y = 4x + b

Now, plug in any of the four points, and solve for b:

7 = 4(1) + b
7 = 4 + b
b = 3

So, now you have your equation:

y = 4x + 3

To be sure you have the right answer, plug in each "x" and make sure that it solves for "y".

In standard form, your equation would be:

4x - y = -3

Answer 4

7