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Here are the two points $(1, 3)$ and $(3, 2)$: What is the formula for the line function whose plot hits both points?

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    Given two points (a,b) and (c,d) what do you know about the line connecting these two points? Have you discussed the point-slope equation of a line in class yet? What have you tried so far? – 2017-02-02
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    Just to be clear, are you asking for the line passing THROUGH these points or are you asking for a line segment CONNECTING the points as end points? – 2017-02-02

5 Answers 5

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Hint

A line function has a form:

$$f(x)=ax+b$$

Plug $(1,3)$ and $(3,2)$ into it and find $a$ and $b$

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Hint: Perhaps try point-slope form:

$$y - y_0 = m(x - x_0)$$

The slope $m$ can be calculated from the two points: $m = (y_1 - y_0)/(x_1 - x_0)$.

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You are given two points in the xy plane. I'm not sure if you have gone over this in class but you need to find the slope of the line formed by these two points. In general, compute y=mx+b, where m is your slope, and b is your y-intercept. The slope formula for this question is y2-y1/x2-x1. Hope this helped a bit.

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​Using y = mx + b will calculate the equation of the line slope = change in y/change in x m = y2 -y1/x2 - x1 m = 2 - (3) / 3 - (1) == -1/2 m = -1/2

y = -1/2x + b y = (-1/2)(1) + b (3) =(-1/2)(1) + b b = 7/2

therefore: y = -1/2x + 7/2​

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    Please use [$\rm \LaTeX$](http://meta.math.stackexchange.com/q/5020/290189). – 2017-02-02
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The equation of a line is denoted as y=mx+b, where m (is your slope) and b (y-intercept).

To find m, we use the following formula, $\frac{y_2-y_1}{x_2-x_1}$. With the given information (1,3) & (3,2) we can see that $x_1=1,y_1=3,x_2=3,y_2=2$.

Now we just simply substitute our values to our slope formula and we get $\frac{2-3}{3-1}$ $\rightarrow -\frac{1}{2}$.

Now we can plug in our slope (m) to the equation of the line and we get, y=$-\frac{1}{2}$x+b.

To get b we simply pick any coordinate point from our given (1,3) or (3,2). Lets pick (1,3) which if you remember the standard notation on a coordinate point is (x,y). So x=1, y=3.

Now let's just plug in to our equation of the line and we get $(3)=-\frac{1}{2}(1)+b.$ Solve for b we get $\rightarrow$ $3=-\frac{1}{2}+b$ (add $\frac{1}{2}$ on both sides)

$\rightarrow \frac{1}{2} +3 = -\frac{1}{2}+\frac{1}{2}+b. \rightarrow \frac{7}{2}=b.$ Then we plug in our last value into the equation of the line and we get:

$y=-\frac{1}{2}+\frac{7}{2}.$