Sketch the plane represents -2x2=-8
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Answer:
Equation of a Line
An important topic of high school algebra is "the equation of a line." This means an equation in x and y whose solution set is a line in the (x,y) plane.
The most popular form in algebra is the "slope-intercept" form
y = mx + b.
This in effect uses x as a parameter and writes y as a function of x: y = f(x) = mx+b. When x = 0, y = b and the point (0,b) is the intersection of the line with the y-axis.
Thinking of a line as a geometrical object and not the graph of a function, it makes sense to treat x and y more evenhandedly. The general equation for a line (normal form) is
ax + by = c,
with the stipulation that at least one of a or b is nonzero. This can easily be converted to slope-intercept form by solving for y:
y = (-a/b) + c/b,
except for the special case b = 0, when the line is parallel to the y-axis.
If the coefficients on the normal form are multiplied by a nonzero constant, the set of solutions is exactly the same, so, for example, all these equations have the same line as solution.
2x + 3 y = 4
4x + 6y = 8
-x - (3/2) y = -2
(1/2)x + (3/4)y = 1
In general, if k is a nonzero constant, then these are equations for the same line, since they have the same solutions.
ax + by = c
(ka)x + (kb)y = kc.
A popular choice for k, in the case when c is not zero, is k = (1/c). Then the equation becomes
(a/c)x + (b/c)y = 1.
Another useful form of the equation is to divide by |(a,b)|, the square root of a2 + b2. This choice will be explained in the Normal Vector section.
Exercise: If O is on the line, show that the equation becomes ax + by = 0, or y = mx.
Exercise: Find the intersections of this line with the coordinate axes.
Exercise: What is the equation of a line through (0,0) and a point (h,k)?
Finding the equation of a line through 2 points in the plane
For any two points P and Q, there is exactly one line PQ through the points. If the coordinates of P and Q are known, then the coefficients a, b, c of an equation for the line can be found by solving a system of linear equations.
Example: For P = (1, 2), Q = (-2, 5), find the equation ax + by = c of line PQ.
Since P is on the line, its coordinates satisfy the equation: a1 + b2 = c, or a + 2b = c.
Since Q is on the line, its coordinates satisfy the equation: a(-2) + b5 = c, or -2 a + 5b = c.
Multiply the first equation by 2 and add to eliminate a from the equation: 4b + 5b = 9b = 2c + c = 3c, so b = (1/3)c. Then substituting into the first equation, a = c - 2b = c - (2/3)c = (1/3)c.
This gives the equation [(1/3)c]x + [(1/3)c}y = c. Why is the c not solved for? Remember that there are an infinite number of equations for the line, each of which is multiple of the other. We can factor out c (or set c = 1 for the same result) and get (1/3)x + (1/3)y =1 as one choice of equation for the line. Another choice might be c = 3: x+y = 3, which has cleared the denominators.
This method always works for any distinct P and Q. There is of course a formula for a, b, c also. This can be found expressed by determinants, or the cross product.
Exercises: Find the equations of these lines. Note the special cases.
Line through (3, 4) and (1, -2).
Line through (3, 4) and (-6, -8).
Line through (3, 4) and (3, 7).