Hints for solving problems in Apply 4.1
(Every time graphing is required, use graph paper)
Graphing Lines I
Problems 1-7
Problem No. 1 - Circle the points below that lie on the line whose equation is
2x - y = 5
Substitute each ordered pair (x, y) into the given equation.
(2, 1) 2(2) - 1 = 5 4 - 1 = 5 False
(3, 1) 2(3) - 1 = 5 6 - 1 = 5 True
(0, -5) 2(0) - (-5) = 5 0 + 5 = 5 True
(-5, 0) 2(-5) - (0) = 5 -10 - 0 = 5 False
(1, -3) 2(1) - (-3) = 5 + 3 = 5 True
So you would circle the ordered pairs: (3, 1), (0, -5), & (1, -3).
Problems 8-28
Problem No. 9 - Graph the equation: x + y = -5
Make a chart: x y Substitute x into the equation and solve for y.
-3 8 -3 + y = 5
-2 7 -2 + y = 5
-1 6 -1 + y = 5
0 5 0 + y = 5
1 4 1 + y = 5
2 3 2 + y = 5
3 2 3 + y = 5
Graph three points: (-1, 6), (0, 5) & (1, 4). Draw the straight line between all three points.
Another way to graph linear equations. (x & y-intercept)
Problem no. 17 - Graph the equation: 2x + 3y = -6.
Make a chart: x y Substitute 0 into the equation.
-3 0 2x + 3(0) = -6 2x = -6 x = -3
0 -2 2(0) + 3y = -6 3y = -6 y = -2
Plot the points (-3, 0) & (0, -2). This is an x-intercept and a y-intercept.
Draw the straight line between the two points. Remember a straight line is determined by two points, but a third point must line on that line.
A third way to graph linear equations: (y = mx + b)
This method is using the slope, y-intercept formula. From the above formula: m is the slope & b is the y-intercept.
Problem No. 15 - Graph the equation: 3x - y = -3.
Solve the problem for y: -y = -3x - 3 y = 3x + 3
From this solved equation for y, you can determine the slope (m), m = 3/1 and the y-intercept (b), b = 3.
So on your graph paper you plot the point (0, 3) and from that point you move one unit to the right and three units up. You will land on the point (1, 6). Draw the line between the two points.
Graphing Lines II
Problems 29-42
Problem No. 29 y = 5
Problem No. 41 x = -5
If the equation has just a y in it (y = b), the graph will be a horizontal line passing through that point on the y axis.
If the equation has just an x in it (x = a), the graph will be a vertical line passing through that point on the x axis.
Problems 43-56
To solve these problem, look above, under another way to graph equations, and you will find out how to determine the x-intercept and the y-intercept.
Slope of a line
Problems 57-58
To solve these problems you use the slope formula:
Problem No. 61 - Find the slope of the line through the points (-2, 7), (4, -5)
(-2, 7) is (x1, y1) and (4, -5) is (x2, y2). Substitute into the formula:
m =
-2 Ans.
Problems 69 & 70
The slope of a horizontal line (y = b) is always 0.
The slope of a vertical line (x = a) is always undefined.
Problems 71-82
Remember parallel lines have the same slope, and perpendicular lines have slopes that are negative reciprocals.
U – Find the slope of a line parallel to the line that passes through the points (12, 2) and (8, -3).
Find the slope of the original line:
So the slope of the original line, given the two points is 5/4, and the slope of the line parallel to the original line is also 5/4.
Problem No. 77 - Find the slope of a line perpendicular to the line that passes through the points (-1, -3) and (4, 7).
Find the slope of the original line:
So the slope of the original line, given the two points is 2, and the slope of the line perpendicular to the original line is the negative reciprocal of 2, which is -1/2.
Problems 83 & 84.
Problem No. 84 - The point (3,2) lies on the line with slope -1/3. Graph this line by finding another point that lies on the line.
Plot the point (3,2) and remember in the slope the numerator is the rise (y-distance) and the denominator is the run (x-distance). Then, from that point go three to the right and one down. You will land on the point (6,1). Now draw the line between the two points and extend it out a little.