20 Analytic geometry in two dimensions
(a) 2x + 3y = 12 (b) 2x - 3y = 12
(c) 2x - 3y = -12 (d) 2x + 3y = -12(e) y=2x+3 (f) Y=0
18 Let k > 0. For each of several values of w, draw the perpendicular
bisector of the line segment joining the points (0, 1/4k) and (w, - 1/4k). Then
determine the condition which x and y must satisfy if the point P(x,y) lies(a) on exactly one of these bisectors,
(b) on more than one of these bisectors,
(c) on none of these bisectors.flns.: (a) y = kx2, (b) y < kx2, (c) y > kx2.19 For what pairs of values of b and c do the two equations3x + by + c = 0
cx - 2y + 12 = 0have the same graph? Partial ans.: There are two pairs which are easily checked
after they have been found.
20 The points PI(xl,yl), P2(xs,y2), P3(x3,ys) are vertices of a triangle. Find
the coordinates (x,y) of the point QI where the line through Pl perpendicular to
the line P2P3 intersects the line P2P3. Partial ans.:(xs - X2)2XI + (ys - yi)(ys - y2)X2 + (y2 - yI)(y2 -ys) x's.
X - (X3 -x2)2 + (ys - y2)2Remark: It is possible to write the answer in different forms. This form enables
us to check quickly that interchanging the subscripts 2 and 3 does not change
the value of x. Such checks are often used to guard against clerical errors in
deriving or copying formulas.
21 Let the vertices of a triangle be Pi(xl,yl), P2(x2,y2), Pa(xa,ys) For each
k, let Lk be the line containing Pk which is perpendicular to the line containing
the other two vertices. Recognizing that altitudes are numbers (not line seg-
ments or lines), we call the lines L1, L2, Ls the altitudinal lines of the triangle.
Prove that these altitudinal lines are concurrent. Remark: The conclusion means
that there is a point Po, called the orthocenter of the triangle, at which the three
altitudinal lines intersect. When asked to prove the conclusion by synthetic
methods, we use our ingenuity (or that of some other people) in searches for
appropriate figures and ideas upon which the proof can be based. When asked
to prove the conclusion by analytic methods, we can proceed at once to apply a
powerful method that cannot fail to produce results if we do the chores correctly
We can find the equations of the three altitudinal lines and use two of the equa-
tions to find the coordinates of the point of intersection of two of the lines. If
this point lies on the third line, the conclusion is true. If (for some triangle)
the point fails to lie on the third line, the conclusion is false. The chores can be
done in the following way. Considering separately the case in which the line
PaPs is neither horizontal nor vertical (so that this line and L1 have slopes) and
the cases in which P,P3 is horizontal or vertical, we can find that the equation