Question

How do you write the point-slope form of the equation below that represents the line that passes through the points (−3, 2) and (2, 1)?

Answer 1

`slope=(y_2-y_1)/(x_2-x_1)`
`y=-1/5x+7/5`

Explanation:

The slope of the line passing through the two points `(-3,2)` and `(2,1)` is given by

`slope=(y_2-y_1)/(x_2-x_1)`

`slope=(1-2)/(2-(-3))`

`slope =-1/5`

The equation of the straight line is

`y=a*x+b`

`slope = a=-1/5`

`y=-1/5x+b`

The point `(2,1)` is a point on the straight line. Plugging the coordinates of this point and the line equation allows us to find the Y-intercept.

`1=-1/5xx2+b`

`b=1+2/5`

`b=7/5`

The equation of the straight-line will be

`y=-1/5x+7/5`

To find the X-intercept assign the value 0 to `y`

`0=-1/5x+7/5`

`1/5x=7/5`

`x=7`

graph{-1/5*x+7/5 [-9.63, 10.37, -3.4, 6.6]}

Answer 2

`y = (-1x)/5 + 7/5`

Explanation:

There are several ways of answering this question:

Method 1 . Use the two points to find the gradient `m`.

`m = (y_2 - y_1)/(x_2 - x_1)" "` Then use one of the points as `(x,y)`

Substitute `m,x and y` into `y = mx + c` and solve to find `c`.

Now use the values for `m and c` in `y = mx + c` to find the required equation.
This method is fine, but it requires several substitutions and it is easy to lose track of where you are.

Method 2 . Use the two points as `x and y` and substitute each into `y = mx + c " "` This forms two equations which can be solved simultaneously to find m and c.

Method 3 . Use the two points as `(x_1,y_1) and (x_2, y_2)` and substitute into the formula for finding the equation of a line if 2 points are known: `" "(y - y_1)/(x - x_1) = (y_2 - y_1)/(x_2 - x_1)`

`" "(y - 1)/(x - 2) = (2 - 1)/(-3 - 2)`

`" "(y - 1)/(x - 2) = 1/-5 " now cross-multiply"`

`-5(y - 1) = (x - 2) " solve for y"`

`-5y + 5 = x - 2`

`-5y " "= x -2 -5`

`y = (-1x)/5 + 7/5 " divide by -5"`