Linear Functions

Definition of Linear Function

A linear function is any function specified by a rule of form $ f: x mx + b $ , where $ m $ . If m = 0, the function is not considered to be a linear function; a function $ f(x) = b $ is called a constant function. The graph of a linear function is always a straight line. The graph of a constant function is a horizontal straight line.

Slope of a Line

The slope of a line that is not parallel to the y-axis is defined as follows (see Figs. 10-1 and 10-2): Let $ (x_{1},y_{1}) $ and $ (x_{2},y_{2}) $ be distinct points on the line. Then the slope of the line is given by

\[ m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{change in y}{change in x}=\frac{rise}{run} \]

  1. Positive slope (line rises) (Fig. 10-1)

Figure 10-1
  1. Negative slope (line falls) (Fig. 10-2)

Figure 10-2

EXAMPLE 10.1 Find the slope of the lines through (a) (5,3) and (8,12); (b) (3,-4) and (-5,6).

  1. Identify $ (x_{1},y_{1})=(5,3) $ and $ (x_{2},y_{2})=(8,12) $ . Then $ m===3 $ .

  2. Identify $ (x_{1},y_{1})=(3,-4) $ and $ (x_{2},y_{2})=(-5,6) $ . Then $ m===- $ .

Horizontal and Vertical Lines

  1. A horizontal line (a line parallel to the x-axis) has slope 0, since any two points on the line have the same y coordinates. A horizontal line has an equation of the form y = k. (See Fig. 10-3.)

  2. A vertical line (a line parallel to the y-axis) has undefined slope, since any two points on the line have the same x coordinates. A vertical line has an equation of the form x = h. (See Fig. 10-4.)

  1. Horizontal line

Figure 10-3
  1. Vertical line

Figure 10-4

Equation of a Line

The equation of a line can be written in several forms. Among the most useful are:

  1. SLOPE-INTERCEPT FORM: The equation of a line with slope m and y intercept b is given by $ y = mx + b $ .

  2. POINT-SLOPE FORM: The equation of a line passing through $ (x_{0}, y_{0}) $ with slope m is given by

\[ y-y_{0}=m(x-x_{0}) \]

  1. STANDARD FORM: The equation of a line can be written as $ Ax + By = C $ , where A, B, C are integers with no common factors; A and B are not both zero.

EXAMPLE 10.2 Find the equation of the line passing through $ (-6,4) $ with slope $ $ .

Use the point-slope form of the equation of a line: $ y - 4 = [x - (-6)] $ . This can then be simplified to slope-intercept form: $ y = x + 8 $ . In standard form, this would become 2x - 3y = -24.

Parallel Lines

If two nonvertical lines are parallel, their slopes are equal. Conversely, if two lines have the same slope, they are parallel; two vertical lines are also parallel.

EXAMPLE 10.3 Find the equation of a line through $ (3,-8) $ parallel to $ 5x + 2y = 7 $ .

First find the slope of the given line by isolating the variable $ y: y = -x + $ . Thus the given line has slope $ - $ . Hence the desired line has slope $ - $ and passes through $ (3, -8) $ . Use the point-slope form to obtain $ y - (-8) = -(x - 3) $ , which is written in standard form as $ 5x + 2y = -1 $ .

Perpendicular Lines

If a line is horizontal, any line perpendicular to it is vertical, and conversely. If two nonvertical lines, with slopes $ m_{1} $ and $ m_{2} $ , are perpendicular, then their slopes satisfy $ m_{1}m_{2} = -1 $ or $ m_{2} = -1/m_{1} $ .

EXAMPLE 10.4 Find the equation of a line through $ (3,-8) $ perpendicular to $ 5x + 2y = 7 $ .

The given line was found in the previous example to have slope $ - $ . Hence the desired line has slope $ $ and passes through $ (3,-8) $ . Use the point-slope form to obtain $ y-(-8)=(x-3) $ , which is written in standard form as 2x-5y=46.

SOLVED PROBLEMS

10.1. For any linear function of form $ f(x) = mx + b $ show that $ = m $ .

Given $ f(x) = mx + b $ , it follows that $ f(x + h) = m(x + h) + b $ , hence

\[ \frac{f(x+h)-f(x)}{h}=\frac{\left[m(x+h)+b\right]-\left[mx+b\right]}{h}=\frac{mx+mh+b-mx-b}{h}=\frac{mh}{h}=m \]

10.2. Which of the following rules represent linear functions?

  1. $ f(x)= $

  2. $ f(x)=x+7 $

  3. $ f(x)=+7 $

Only (b) represents a linear function. The rule in (a) represents a constant function, while the rule in (c) is referred to as a nonlinear function.

10.3. Find the equation of the horizontal line through $ (5,-3) $ .

A horizontal line has an equation of the form \(y = k\). In this case, the constant \(k\) must be \(-3\). Hence the required equation is \(y = -3\).

10.4. Find the equation of the vertical line through $ (5,-3) $ .

A vertical line has an equation of the form x = h. In this case, the constant h must be 5. Hence the required equation is x = 5.

10.5. Find the equation of the line through $ (-6,8) $ with slope $ $ . Write the answer in slope-intercept form and also in standard form.

Use the point-slope form of the equation of a line, with $ m = $ and $ (x_{0}, y_{0}) = (-6, 8) $ . Then the equation of the line can be written:

\[ y-8=\frac{3}{4}[x-(-6)] \]

Simplifying, this becomes $ y = x + $ in slope-intercept form, and $ -3x + 4y = 50 $ in standard form.

10.6. Find the equation of the line through the points $ (3,-4) $ and $ (-7,2) $ . Write the answer in slope-intercept form and also in standard form.

First, find the slope of the line: Identify \((x_{1},y_{1})=(3,-4)\) and \((x_{2},y_{2})=(-7,2)\). Then

\[ m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2-(-4)}{-7-3}=-\frac{3}{5} \]

Now, use the point-slope form of the equation of a line, with $ m = - $ . Choose either of the given points, say, $ (3, -4) = (x_{0}, y_{0}) $ . Then the equation of the line can be written:

\[ y-(-4)=-\frac{3}{5}(x-3) \]

Simplifying, this becomes $ y = -x - $ in slope-intercept form and $ 3x + 5y = -11 $ in standard form.

10.7. (a) Show that the equation of a line with x-intercept a and y-intercept b, where neither a nor b is 0, can be written as $ + = 1 $ . (This is known as the two-intercept form of the equation of a line.) (b) Write the equation of the line with x-intercept 5 and y-intercept -6 in standard form.

  1. The line passes through the points $ (a,0) $ and $ (0,b) $ . Hence its slope can be found from the definition of slope as $ m===- $ . Using the slope-intercept form, the equation of the line can be written as $ y=-x+b $ , or $ x+y=b $ . Dividing by b yields $ +=1 $ as required.

  2. Using the result of part (a) gives $ + = 1 $ . Clearing of fractions yields 6x - 5y = 30 in standard form as required.

10.8. It is shown in calculus that the slope of the line drawn tangent to the parabola $ y = x^{2} $ at the point $ (a, a^{2}) $ has slope 2a. Find the equation of the line tangent to $ y = x^{2} $ (a) at (3, 9); (b) at $ (a, a^{2}) $ .

  1. Since the line has slope $ 2 = 6 $ , use the point-slope form to find the equation of a line through $ (3,9) $ with slope 6: $ y - 9 = 6(x - 3) $ or y = 6x - 9.

  2. Since the line has slope 2a, use the point-slope form to find the equation of a line through $ (a, a^{2}) $ with slope 2a: $ y - a^{2} = 2a(x - a) $ or $ y = 2ax - a^{2} $ .

10.9. Prove that two nonvertical lines are parallel if and only if they have the same slope. (See Fig. 10-5.)

Let $ l_{1} $ and $ l_{2} $ be two different lines with slopes, respectively, $ m_{1} $ and $ m_{2} $ , and y-intercepts, respectively, $ b_{1} $ and $ b_{2} $ .

Figure 10-5

Then the lines have equations $ y = m_{1}x + b_{1} $ and $ y = m_{2}x + b_{2} $ . The lines will intersect at some point $ (x, y) $ if and only if for some x the values of y are equal, that is,

\[ m_{1}x+b_{1}=m_{2}x+b_{2};thus,(m_{1}-m_{2})x=b_{2}-b_{1} \]

This is possible, that is, the lines intersect, if and only if $ m_{1} m_{2} $ . Hence the lines are parallel if and only if $ m_{1} = m_{2} $ .

10.10. Find the equation of the line through $ (5,-3) $ parallel to (a) y = 3x - 5; (b) $ 2x + 7y = 4 $ ; (c) x = -1.

  1. Any line parallel to the given line will have the same slope as the given line. Since the given line is written in slope-intercept form, its slope is clearly seen to be 3. The equation of a line through $ (5,-3) $ with slope 3 is found from the point-slope form to be $ y-(-3)=3(x-5) $ . Simplifying yields y=3x-18.

  2. It is possible to proceed as in (a); however, the equation of the given line must be analyzed to find its slope. An alternative method is to note that any line parallel to the given line can be written as $ 2x + 7y = C $ . Then, since $ (5, -3) $ must satisfy the equation, $ 2 + 7(-3) = C $ ; hence C = -11 and $ 2x + 7y = -11 $ is the required equation.

  3. Proceeding as in (b), note that any line parallel to the given line must be vertical, hence must have an equation of the form x = h. In this case, h = 5; hence x = 5 is the required equation.

10.11. Prove that if two lines with slopes m and m, are perpendicular, then m1m2 = −1. (Fig. 10-6.)

The slopes of the lines must have opposite signs. In the figure, $ m_{1} $ is chosen (arbitrarily) positive and $ m_{2} $ is chosen negative.

Figure 10-6

Since $ l_{1} $ has slope $ m_{1} $ , a run of 1 (segment PB) yields a rise of $ m_{1} $ along $ l_{1} $ (segment CB). Similarly, since $ l_{2} $ has slope $ m_{2} $ , a run of 1 yields a (negative) rise of $ m_{2} $ along $ l_{2} $ , thus segment AB has length $ -m_{2} $ . Since the lines are perpendicular, the triangles PCB and APB are similar. Hence ratios of corresponding sides are equal; it follows that

\[ \frac{CB}{PB}=\frac{PB}{AB} \]

\[ \frac{m_{1}}{1}=\frac{1}{-m_{2}} \]

\[ m_{1}m_{2}=-1 \]

10.12. Find the equation of the line through $ (8,-2) $ perpendicular to (a) $ y=x+2 $ ; (b) $ x+3y=6 $ ; (c) x=7.

  1. Any line perpendicular to the given line will have slope m satisfying $ m = -1 $ ; thus, $ m = - $ . The equation of a line through $ (8, -2) $ with slope $ - $ is found from the point-slope form to be $ y - (-2) = -(x - 8) $ . Simplifying yields $ 5x + 4y = 32 $ .

  2. First, determine the slope of the given line. Isolating the variable y, the equation is seen to be equivalent to $ y = -x + 2 $ ; hence the slope is $ - $ . Any line perpendicular to the given line will have slope m satisfying $ -m = -1 $ ; thus, m = 3. The equation of a line through $ (8, -2) $ with slope 3 is found from the point-slope form to be $ y - (-2) = 3(x - 8) $ . Simplifying yields y = 3x - 26.

  3. Since the given line is vertical, any line perpendicular to the given line must be horizontal, hence must have an equation of the form y = k. In this case, k = -2; hence y = -2 is the required equation.

10.13. Find the rule for a linear function, given $ f(0) = 5 $ and $ f(10) = 12 $ .

Since the graph of a linear function is a straight line, this is equivalent to finding the equation of a line in slope-intercept form, given a y-intercept of 5. Since the line passes through $ (0,5) $ and $ (10,12) $ , the slope is determined; $ m = = $ . Hence the equation of the line is $ y = x + 5 $ and the rule for the function is $ f(x) = x + 5 $ .

10.14. Find a general expression for the rule for a linear function, given $ f(a) $ and $ f(b) $ .

Since the graph of a linear function is a straight line, this is equivalent to finding the equation, in slope-intercept form, of a line passing through $ (a, f(a)) $ and $ (b, f(b)) $ . Clearly, a line passing through these two points will have slope $ m = $ . From the point-slope form, the equation of the line can be written as $ y - f(a) = (x - a) $ or $ y = (x - a) + f(a) $ . Thus the rule for the function is

\[ f(x)=\frac{f(b)-f(a)}{b-a}(x-a)+f(a) \]

10.15. Find the rule for a linear function, given $ f(10) = 25,000 $ and $ f(25) = 10,000 $ .

Apply the formula from the previous problem with \(a = 10\) and \(b = 25\). Then

\[ \begin{aligned}f(x)&=\frac{10,000-25,000}{25-10}(x-10)+25,000\\&=-1000x+35,000\end{aligned} \]

10.16. Suppose the cost of producing 50 units of a given commodity is 27,000, while the cost of producing 100 units of the same commodity is 38,000. If the cost function $ C(x) $ is assumed to be linear, find a rule for $ C(x) $ . Use the rule to estimate the cost of producing 80 units of the commodity.

This is equivalent to finding the equation of a straight line passing through $ (50,27000) $ and $ (100,38000) $ . The slope of this line is $ m==220 $ ; hence from the point-slope form the equation of the line is

$ y - 27,000 = 220(x - 50) $ . Simplifying yields $ y = 220x + 16,000 $ ; thus, the rule for the function is $ C(x) = 220x + 16,000 $ .

The cost of producing 80 units of the commodity is given by $ C(80) = 220 + 16,000 $ or 33,600.

10.17. If the value of a piece of equipment is depreciated linearly over a 20-year period, the value $ V(t) $ can be described as a linear function of time t.

  1. Find a rule for $ V(t) $ assuming that the value at time t = 0 is $ V_{0} $ and that the value after 20 years is zero.

  2. Use the rule to find the value after 12 years of a piece of equipment originally valued at $7500.

  3. This is equivalent to finding the equation of a line of form $ V = mt + b $ , where the V-intercept is $ b = V_{0} $ , and the line passes through $ (0, V_{0}) $ and $ (20, 0) $ . The slope is given by $ m = = - $ ; hence the equation is $ V = -t + V_{0} $ and the rule for the function is $ V(t) = -t + V_{0} $ .

  4. In this case, $ V_{0} = 7500 $ and the value of $ V(12) $ is required. Since the rule for the function is now $ V(t) = -t + 7500 = 7500 - 375t $ , $ V(12) = 7500 - 375 = 3000 $ and the value is 3000.

SUPPLEMENTARY PROBLEMS

10.18. Write the following equations in standard form:

\[ \left(a\right)y=3x-2;\left(b\right)y=-\frac{1}{2}x+8;\left(c\right)y=\frac{2}{3}x-\frac{3}{5} \]

Ans. (a) 3x - y = 2; (b) $ x + 2y = 16 $ ; (c) 10x - 15y = 9

10.19. Write the following equations in slope-intercept form:

\[ Ans.\quad(a)y=-\frac{1}{3}x+\frac{7}{6};(b)y=\frac{3}{5}x-3;(c)y=-\frac{3}{4}x+\frac{9}{8} \]

10.20. Find the equation of a line in standard form given:

  1. The line is horizontal and passes through $ (,) $ .

  2. The line has slope -0.3 and passes through $ (1.3, -5.6) $ .

  3. The line has x-intercept 7 and slope -4.

  4. The line is parallel to y = 3 - 2x and passes through the origin.

  5. The line is perpendicular to 3x - 5y = 7 and passes through $ (-,) $ .

  6. The line passes through $ (a,b) $ and $ (c,d) $ .

Ans. (a) 4y = 3

\[ 30x+100y=-521 \]

  1. $ 4x + y = 28 $

\[ 2x+y=0 \]

  1. $ 10x + 6y = -9 $

  2. $ (b - d)x + (c - a)y = (b - d)a + (c - a)b $

10.21. Find the equation of a line in slope-intercept form given:

  1. The line is horizontal and passes through $ (-3,8) $ .

  2. The line has slope $ - $ and passes through $ (-5,1) $ .

  3. The line has x-intercept -2 and slope $ $ .

  4. The line is parallel to $ 2x + 5y = 1 $ and passes through $ (2, -8) $ .

  5. The line is perpendicular to $ y = x - 1 $ and passes through $ (6,0) $ .

Ans. (a) y = 8; (b) $ y = -x - $ ;

  1. $ y = x + 1; $

  2. $ y = -x - ; $

  3. $ y = -x + 16 $

10.22. Find the slope and the y-intercept for (a) y = 5 - 3x; (b) $ 2x + 6y = 9 $ ; (c) $ x + 5 = 0 $ .

Ans. (a) slope -3, y-intercept 5; (b) slope $ - $ , y-intercept $ $ ; (c) slope undefined, no y-intercept

10.23. Find the possible slopes of a line that passes through $ (4,3) $ so that the portion of the line in the first quadrant forms a triangle of area 27 with the positive coordinate axes.

\[ Ans.\quad-\frac{3}{2}or-\frac{3}{8} \]

10.24. Repeat problem 10.23 except with the triangle having area 24.

Ans. $ - $ is the only possible slope.

10.25. Recall from geometry that the line drawn tangent to a circle is perpendicular to the radius line drawn to the point of tangency. Use this fact to find the equation of the line tangent to

  1. the circle $ x^{2} + y^{2} = 25 $ at $ (-3,4) $

  2. the circle $ (x - 2)^{2} + (y + 4)^{2} = 4 $ at $ (2, -2) $

Ans. (a) 3x - 4y = -25; (b) y = -2

10.26. It is shown in calculus that the slope of the line drawn tangent to the curve $ y = x^{3} $ at the point $ (a, a^{3}) $ has slope $ 3a^{2} $ . Find the equation of the line tangent to $ y = x^{3} $ at (a) (2, 8); (b) $ (a, a^{3}) $ .

\[ Ans.\quad(a)y=12x-16;(b)y=3a^{2}x-2a^{3} \]

10.27. The line drawn perpendicular to the tangent line to a curve at the point of tangency is called the normal line. Find the equation of the normal line to $ y = x^{3} $ at (2,8). (See the previous problem.)

Ans. $ y = -x + $

10.28. An altitude of a triangle is a line drawn from a vertex of the triangle perpendicular to the opposite side of the triangle. Find the equation of the altitude drawn from $ A(0,0) $ to the side formed by $ B(3,4) $ and $ C(5,-2) $ .

\[ Ans.\quad x-3y=0 \]

10.29. A median of a triangle is a line drawn from a vertex of the triangle to the midpoint of the opposite side of the triangle. Find the equation of the median drawn from $ A(5,-2) $ to the side formed by $ B(-3,9) $ and $ C(4,-7) $ .

\[ Ans.\quad2x+3y=4 \]

10.30. Find a rule for a linear function given $ f(5) = -7 $ and $ f(-5) = 10 $ .

Ans. $ f(x) = -x + $

10.31. Find a rule for a linear function given $ f(0) = a $ and $ f(c) = b $ .

Ans. $ f(x)=x+a $

10.32. In depreciation situations (see Problem 10.17), it is common that a piece of equipment has a residual value after it has been linearly depreciated over its entire lifetime.

  1. Find a rule for $ V(t) $ , the value of a piece of equipment, assuming that the value at time t = 0 is $ V_{0} $ and that the value after 20 years is R.

  2. Use the rule to find the value after 12 years of a piece of equipment originally valued at $7500, assuming that it has a residual value after 20 years of $500.

Ans. (a) $ V(t) = t + V_{0} $ ; (b) 3300