


Linear relations and their graphing
LINEAR RELATIONS
LINEAR RELATION A linear relation in two variables is a relation that can be written in the form y=ax+b, In the equation Ax+By=C, any number can be used for x or y, so both the domain and range of a linear relation in which neither Anor B is 0are the set of real numbers (− ∞,∞), GRAPHING LINEAR RELATIONS. The graph of a linear relation can be found by plotting at least two points. Two points that are especially useful for sketching the graph of a line are found with the intercepts. An xintercept is an xvalue at which a graph crosses the xaxis. A yintercept is a yvalue at which a graph crosses the yaxis. Since y = 0 on the xaxis, an xintercept is found by setting y equal to 0 in the equation and solving for x. Similarly, a yintercept is found by settingx=0 in the equation and solving for y. Example 1 GRAPHING A LINEAR RELATION USING INTERCEPTS Graph 3 x+2 y=6. 2 y=6 y=3
Let’s see how our math solver generates graphs of this and similar problems. Click on "Solve Similar" button to see more examples. Example 2 GRAPHING HORIZONTAL AND VERTICAL LINES (a) Graph y=− 3. Since y always equals − 3, the value of y can never be 0. This means that the graph has no xintercept. The only way a straight line can have no xintercept is for it to be parallel to the xaxis, as shown in Figure 3.8. Notice that the domain of this linear relation is (− ∞,∞) but the range is {− 3}.
(b) Graph x=− 3. Here, since x always equals − 3, the value of x can never be 0, and the graph has no yintercept. Using reasoning similar to that of part (a), we find that this graph is parallel to the yaxis, as shown in Figure 3.9. The domain of this relation is {− 3}, while the range is (− ∞,∞), From this example we may conclude that a linear relation of the form y=k has as its graph a horizontal line through (0,k), and one of the form x=k has as its graph a vertical line through (k,0).
Example 3 GRAPHING A LINE THROUGH THE ORIGIN Graph 4 x5 y=0, Find the intercepts. If x=0, then Letting y=0 leads to the same ordered pair,0=0. The graph of this relation has just one intercept—at the origin. Find another point by choosing a different value for x (or y). Choosing x=5 gives 205 y=0 20=5 y 4=y
Let’s see various graphs of line passing through origin. Click on "Solve Similar" button to see more examples. SLOPE An important characteristic of a straight line is its slope, a numerical measure of the steepness of the line. (Geometrically, this may be interpreted as the ratio of rise to run.) To find this measure, start with the line through the two distinct points (x_{1},y_{1}) and (x_{2},y_{2}), as shown in Figure 3.11 , where x_{1}≠x_{2}.The difference
is called the change in x and denoted by Δ (x) (read “delta x’), where Δ is the Greek letter delta. In the same way, the change in y can be written Δ y=y_{2}y_{1} The slope of a nonvertical line is defined as the quotient of the change in y and the change in x, as follows. SLOPE The slope m of the line through the points (x_{1},y_{1}) and (x_{2},y_{2}) is m=Δ yΔ x=y_{2}y_{1}x_{2}x_{1} Where Δ (x)≠0 CAUTION When using the slope formula, be sure that it is applied correctly. It makes no difference which point is (x_{1},y_{1}) or (x_{2},y_{2}); however, it is important to be consistent. Start with the x and yvalue of one point (either one) and subtract the corresponding values of the other point. The slope of a line can be found only if the line is nonvertical. This guarantees that x_{2}≠x_{1}, so that the denominator (x_{2}x_{1})≠0. It is not possible to define the slope of a vertical line. The slope of a vertical line is undefined. Example 4.FINDING SLOPES WITH THE SLOPE FORMULA Find the slope of the line through each of the following pairs of points. (a) (− 4,8),(2,− 3) Δ y=− 38=− 11 and x Δ=2(− 4)=6 The Slope is m=Δ yΔ x=− 116 (b) (2,7),(2,− 4) A sketch would show that the line through (2,7) and (2,− 4) is vertical. As mentioned above, the slope of a vertical line is not defined. (An attempt to use the (c) (5,− 3) and (− 2,− 3) By definition of slope, Drawing a graph through the points in Example 4(c) would produce a line that is horizontal, which suggests the following generalization. The slope of a horizontal line is 0. Figure 3.12 shows lines of various slopes. As the figure shows, a line with a positive slope goes up from left to right, but a line with a Positive slope negative slope goes down from left to right. It can be shown, using theorems for similar triangles, that the slope slope is independent of the choice of points on the line. That is, the slope of a line is the same no matter which pair of distinct points on the line are used to find it. Since the slope of a line is the ratio of vertical change to horizontal change, if we know the slope of a line and the coordinates of a point on the line, the graph of the line can be drawn. The next example illustrates this.
Example 5 GRAPHING A LINE USING A POINT AND THE SLOPE Graph the line passing through (− 1,5) and having slope− 53. This gives a second point, (2,0), which can then be used to complete the graph. Because − 53=5− 3, another point could be obtained by starting at (− 1,5)and moving 5 units up and 3 units to the left. We would reach a different second point, but the line would be the same.
EQUATIONS OF A LINE Since equations can define relations, we now consider methods of finding equations of linear relations. Figure 3.14 shows the line passing through the fixed point (x_{1},y_{1}) and having slope m. (Assuming that the
This result, called the pointslope form of the equation of a line, identifies points on a given line: a point (x,y) lies on the line through (x_{1},y_{1}) with slope m if and only if yy_{1}=m (xx_{1}) the pointslope form of the equation of a line. Write an equation of the line through (− 4,1) with slope − 3. Here x_{1}=− 4, y_{1}=1, and m=− 3. Use the pointslope form of the equation of a line to get y1=− 3 [x(− 4)] x_{1}=− 4y_{1}=1 m=− 3 y1=− 3 (x+4) y1=− 3 x12 Distributive property or 3 x+y=− 11 in standard form. CAUTION The definition of “standard form” is not standard from one text to another. Any linear equation can be written in many different (all equally correct) forms. For example, the equation 2 x+3 y=8 can be written as 2 x=83 y , 3 y=82 x ,x+32 y=4,4 x+6 y=16 and so on. In addition to writing it in the form Ax+By=C(with A≥0), let us agree that the form 2 x+3 y=8 is preferred over any multiples of both sides, such as 4 x+6 y=16. Example 7 USING THE POINTSLOPE FORM (GIVEN TWO POINTS) Find an equation of the line through (− 3,2) and (2,− 4)
6 x+5 y=− 8 Standard form Verify that the same equation results if (2,− 4) is used instead of (− 3,2) in the Pointslope form. As a special case of the pointslope form of the equation of a line, suppose that a line passes through the point (0,b), so the line has yintercept b. If the line has slope m, then using the pointslope form with x_{1}=0 and y_{1}=b gives yb=m (x0) y=mx+b SLOPEINTERCEPT FORM The line with slope m and yintercept b has an equation Find the slope and yintercept of 3 xy=2 Graph the line using this information. First write 3 xy=2 in the slopeintercept form, y=mx+b, by solving for y, getting 3 xy=2. This result shows that the Slope is m=3 and the yintercept is b=− 2. To draw the graph, first locate the yintercept. See Figure 3.15. Then, as in Example 5, use the slope of 3, or 31, to get a second point on the graph. The line through these two points is the graph of 3 xy=2.
In the preceding discussion, it was assumed that the given line had a slope. The only lines having undefined slope are vertical lines. The vertical line through the point (a,b)passes through all the points of the form (a,y), for any value of y. This fact determines the equation of a vertical line. The horizontal line through the point (a,b) passes through all points of the For example, the horizontal line through (1,− 3) has the equation y=− 3. See Figure 3.8 for the graph of this equation. The equation of the xaxis is y=0. PARALLEL. AND PERPENDICULAR LINES Slopes can be used to decide whether or not two lines are parallel. Since two parallel lines are equally “steep,” they should have the same slope. Also, two distinct lines with the same “steepness” are parallel. The following result summarizes this discussion. PARALLEL LINES Two distinct non vertical lines are parallel if and only if they have the same slope. For example, if the slope of a line is − 34, the slope of any line perpendicular Example 9 USING THE SLOPE RELATIONSHIPS FOR PARALLEL AND PERPENDICULAR Find the equation of the line that passes through the point (3,5) and satisfies the given condition. (a) parallel to the line 2 x+5 y=4 y=− 25 x+45 yy_{1}=m (xx_{1}) y5=− 25 (x3) 5 (y5)=− 2 (x3) (5 y25)=− 2 x+6 2 x+5 y=31 (b) perpendicular to the line 2 x+5 y=4 2 (y5)=5 (x30) 2 y10=5 x15 − 5 x+2 y=− 5 or 5 x2 y=5 All the lines discussed above have equations that could be written in the form Ax+By=C for real numbers A, B, and C. As mentioned earlier, the equation Ax+By=C is the standard form of the equation of a line. The various forms of linear equations are listed below.LINER EQUATIONS
PROBLEM SOLVING A straight line is often the best approximation of a set of data points that result from a real situation. If the equation is known, it can be used to predict the value of one variable, given a value of the other. For this reason, the equation is written as a linear relation in slopeintercept form. One way to find the equation of such a straight line is to use two typical data points and the pointslope form of the equation of a line. Example 10 FINDING AN EQUATION FROM DATA POINTS Scientists have found that the number of chirps made by a cricket of a particular Species per minute is almost linearly related to the temperature. Suppose that for a particular species, at 68°F a cricket chirps 124 times per minute, while at 80° F the cricket chirps 172 times per minute. Find the linear equation that relates the number of chirps to the temperature. Think of the ordered pairs in the relation as (chirps, temperature), or (c,t). Then c takes on the role of x and t takes on the role of y. Since we are using a linear relationship, find the slope of the line by using the slope formula with the points (124,68) and (172,80). t68=14 c31
