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In this lesson, you will learn a formula to compute the slope of a line using any two points on the line (not just points whose $x$-coordinates differ by 1). You will also use similar triangles to see why this formula works, and why the line with slope $m$ and $y$-intercept $(0, b)$ is the graph of the equation $y=mx+b$.
Jane is descending a 20-foot rock. She climbs down the rock at a rate of 2 feet per minute (so, each minute her height decreases by 2 feet).
The rate of change between the points $(0, 20)$ and $(2, 16)$ is:
Find the rate of change between the points shown in the table below. All of these points are on the graph.
The slope of the line through the points $(x_1,y_1)$ and $(x_2,y_2)$ is: $${y_2-y_1}/{x_2-x_1}$$
The formula $${y_2-y_1}/{x_2-x_1}$$ from slopeDef lets you compute slopes by using $x$-values that aren’t exactly one unit apart. Sometimes, this makes it easier to compute those slopes.
Compute the slope of each line by using the given $x$-values, and the formula $${y_2-y_1}/{x_2-x_1}$$.
Compute the slope of the line $$y=-3/2x+1$$ by using each of the given $x$-values.
As you can see, whenever the $x$-coordinate on the graph of $$y=-3/2x+1$$ changes by 2, the $y$-coordinate changes by $-3$ (goes down by 3). This is illustrated by the picture below.
You can say something similar for any line whose slope is a fraction:
For any numbers $p$ and $q$ with $q$ not equal to zero, a line of slope $$p/q$$ is the graph of an equation of the form $$y=p/q x + b$$. Changing the $x$-coordinate on that line by $q$ makes the $y$-coordinate change by $p$.
You have seen that it’s possible to compute the slope of a line by using any two points on that line, and you will get the same answer whichever points you choose. We will now explore why that is true.
Compute the slope of each line in the table below, by using the given $x$-values. In each case, a triangle will be drawn on the grid to the left, to the right of the line that is graphed. The horizontal side of the triangle shows the difference between the two $x$-values in the table, while the vertical side shows the difference between the two $y$-values. So the lengths of the sides of the triangle determine the numerator ($y_2-y_1$) and denominator ($x_2-x_1$) of the slope computation $${y_2-y_1}/{x_2-x_1}$$.
When the slope of a line is negative, the vertical side of the triangle will be below the horizontal side instead of above it. (Its length will still be positive, because lengths are always positive.)
The line $$\cl"red"{y = 1/2 x + 2}$$ is graphed on the grid to the left, along with a triangle like the ones in slope-triangle-qn, representing a slope computation for this line. You can slide the $d$ slider to translate (move) the triangle along the line, and slide the $r$ slider to dilate (expand and contract) the triangle about its bottom left point.
This is because translations and dilations take lines to parallel lines, so they transform horizontal lines to other horizontal lines, and vertical lines to other vertical lines. Also, these translations and dilations don’t change the red line.
This works because translations don’t change length measurements (they are rigid motions).
Dilations change length measurements, but they multiply all lengths by the same number $r$. So dilations don’t change the result of the slope computation $${y_2-y_1}/{x_2-x_1}$$, which depends on the quotient (or ratio) of two length measurements.
Any two triangles that you get in this way can be transformed into each other by translating and dilating, so the triangles are similar (have the same shape, but not necessarily the same size).
The line $$\cl"red"{y=3/4 x - 1}$$ is graphed on the grid to the left. Two triangles like the ones in slope-triangle-qn are drawn on that line, representing two different slope computations for this line. You can slide the $d$ slider to translate (move) one of those triangles along the line, and slide the $r$ slider to dilate (expand and contract) that triangle about its bottom left point. The transformed triangle will be drawn in red.
Because any two slope computations for this line can be connected by these transformations, and these transformations do not change the value you get from a slope computation, any two slope computations must result in the same value. That is, you can use the formula $${y_2-y_1}/{x_2-x_1}$$ with any two points $(x_1,y_1)$ and $(x_2,y_2)$ on the line, and you will get the same number (the slope of the line). This is true for any non-vertical line.
The line drawn to the left passes through the origin $\cl"red"{(0,0)}$, and has slope 2. You can slide the $x$ slider to see various slope computation triangles for the line, using $(x_1,y_1) = \cl"red"{(0,0)}$ and $(x_2,y_2) = \cl"red"{(x,y)}$ for the point $\cl"red"{(x,y)}$ on the line.
Click to see a line through the origin with slope $-2$. You can slide the $x$ slider to see various slope computation triangles for the line, using the points $(x_1,y_1) = \cl"red"{(0,0)}$ and $(x_2,y_2) = \cl"red"{(x,y)}$ on the line.
Click to add an $m$ slider below the grid to the left, which controls the line’s slope. A slope computation triangle is still shown, using the points $(x_1,y_1) = \cl"red"{(0,0)}$ and $(x_2,y_2) = \cl"red"{(x,y)}$ on the line.
As you can see:
If a line passing through the origin has slope $m$, its equation is $y=mx$.
The line drawn to the left has slope $2$ and $y$-intercept $\cl"red"{(0,1)}$. You can slide the slider to see various slope computation triangles for the line, along with the distance (marked in blue) between the horizontal side of each slope computation triangle and the $x$-axis.
For each row in the table below, give an equation for the line with the given slope and $y$-intercept.
If a line has slope $m$ and $y$-intercept $(0, b)$, it is the graph of the equation $y=mx+b$.