|UBC Calculus Online Course Notes|
Equations of Straight Lines
This page is a quick review of equations of straight lines. It contains a summary of the different ways of arriving at the equation of a straight line. Consult your favourite Calculus book for more details. As mentioned in the discussion of lines on the previous page, only two pieces of information are needed to completely describe a given line. However, we have some flexibility on which two pieces of information we use
(Comment: Since the "y intercept" is simply the point (0,b), the first and second entries above are closely related.)
Example 1 Find the equation of a straight line through the point (-1,3) with slope 2.
Answer: The slope of the line must be the same between any two points, e.g. between (x,y) and (-1,3) so that
This implies that
Find the equation of a line through the points (1,2) and (3,1). What is its slope? What is its y intercept?
Answer: We first find the slope of the line by finding the ratio of the change in y over the change is x. Thus
Now that we have the slope, the problem is similar to that in Example 1, so we use the same method to conclude that
Rearranging leads to
Thus, the y intercept is 5/2 and the slope is -1/2.
What is the equation of the line that passes through the point (1,1) and is perpendicular to the line ? Where do the two lines intersect ?
Recall that two lines are perpendicular if their slopes are negative reciprocals of one another. Thus, the line would be perpendicular to any line whose slope is .
The slope of the original line is -2. Thus the slope of a second line perpendicular to it would be 1/2. We also require that this second line go through the point (1,1). Now we have reduced the problem to the type of calculation that we already did in example 1. We find that points on the line satisfy
After rearranging, we find that the required equation is:
In the second part of this problem we are asked to find the point(s) at which the two lines intersect. First recall that two lines either intersect at exactly one point, or else they intersect all along their length (if they correspond to the same line !). In any case, at points of intersection, both equations are satisfied simultaneously. Thus, we look for points (x,y) such that
The above two equations imply that
Thus, the x-coordinate of the point of intersection is found. The y value can be obtained by using either of the two equations, and simply plugging in this value of x. We find that
The answer is that the point of intersection is . It is wise to check that this point also satisfies the equation of the second line to avoid simple algebraic mistakes.
Find the equation of the tangent line to the circle
at the point in the first quadrant whose x coordinate is x=1.
In order to answer this question, we must review some basic ideas from geometry. First, recall that a tangent line has the property that it "touches" a curve, intersecting it at a single point and having "the same direction" as the curve. (This vague description will be made much more precise in the course of this Calculus class). The special property of a circle is that the tangent line at each point on the circle is perpendicular to the radius vector. (A radius vector is a line pointing from the center of the circle to the point on the rim of the circle). You can see these properties of the tangent line in the demonstration below. (Move the red dot with your mouse to see how the tangent line, shown in red changes at different positions on the circle.)
The equation given in this problem describes a circle with center at (0,0) and radius 2. Further, note that if the x coordinate of the point on the circle is x=1, then the y coordinate must be given by
We observe that there are two values of y that correspond to the x coordinate x=1, but since we are interested in a point in the first quadrant, we select the positive value. This means that the point of the circle at which we want a tangent line has the coordinates . Thus, the radius vector is a line connecting (0,0) and , which means that its slope is
The tangent line, by what we have said above, is perpendicular to this radius vector, and has a slope the negative reciprocal of the above, namely
We now have all the information required to compute the equation of the tangent line: namely its slope, and a point through which it passes, . We find, after some simple algebra that the equation of the tangent line is: