2

QB=√x12+(mx1)2=x1√1+m2.

The ▲ PMQ is similar to the ▲ BNQ; hence

When the point (x, y) is on one side of BM, y1-mx1-b is negative; when (x, y) is on the other side, y1-mx,-b is positive; when (x1, y1) is on BM, y1—mî—b is zero. Only the numerical value of d is important.

1

The distance from (x, y) to Ax+By+C=0, or to

For instance, required the distance from (4, -3) to

Examples.

1. Find the equation of the mid-perpendicular to the line joining (1, 3) and (3,

5).

2. Find the following distances: From (-3, 7) to (0,3); from (-2,-5) to (3,7). Ans. 5 and 13.

3. The vertices of a triangle being A(−2,0), B(2,0), and C(1, 5), find the middle point of each side, and the lengths of the lines joining the mid-points. Check by showing them to be one-half as long as the sides.

4. Find the equations of AC and BC in Example 3 and the lengths of the altitudes of the triangle perpendicular to them. 5. Find the distance of (-2, 1) from 3x-4y=5. Ans. 3. 6. Find the distance of (1, 3) from 12y=5x+11.

107. Intersections of Lines; Simultaneous Solutions.-Suppose we have given the equations of two lines:

AB: 3x-y+2=0,

CD: x+2y-6=0,

and wish to find the coordinates of their common point, P. (See Fig. 16.)

The coordinates of P satisfy the equation of AB, because P is on AB; and they satisfy the equation of CD, because P is on CD. As they thus satisfy both equations, they can be obtained by solving the two equations simultaneously. Solving, we get x=4, y=2, so that P is the point (7, 24).

This figure may evidently be used to represent graphically the common solution of the two simultaneous equations.

In general:

The coordinates of any common point of two curves may be found by the simultaneous solution of the equations of the curves, and:

Any pair of values of x and y satisfying two simultaneous equations in x and y are the x and y coordinates of a point common to the graphs of the equations.

108. A special case is encountered if we attempt to solve two such equations as

2x-3y=7,
2x-3y=8;

for any attempt at elimination of one variable, leads to the elimination of both, with an absurd result, 1-0. We have seen that such a result indicates that the root to be expected has disappeared through becoming infinite. This is consistent, as the two lines are parallel, and form the limiting case of lines meeting at an increasingly great distance from the origin. For instance,

2x 3.01y and 2x-3y=8 meet at (154, 100), 2x-3.0001y 7 and 2x-3y=8 meet at (15004, 10000),

and so on.

This relation is expressed by saying that parallel lines meet at infinity.

This appears also in another way. If (x, y) is any point on

Y1

the line y=mx+b, the ratio of the coordinates is

X1

and as x (and therefore y1 also) increases indefinitely,

the ratio of y to x at co thus being independent of b.

Examples.

Show graphically and estimate the simultaneous solutions of the following pairs of equations:

1. 3y-2x=5 and 6y-x=3.

2. 5y+12x=8 and 4y-10x=7.

3. 5y-12x-4 and 3y-7x=6. (Use a small scale.)

109. Finding Geometric Loci by Algebraic Processes.-To find the locus of points equidistant from two given points, A and B. Call the distance AB=2a; take AB as x-axis and the mid-perpendicular to AB as y-axis. Then (see Fig. 17) the coordinates of A are (—a, 0); of B, (a, 0). Let the coordinates

of any point P of the desired locus be (x, y). Then, by hypothesis, the distance PA=the distance PB, or

√[x-(-a)]2+(y−0)2=√(x−a)2+(y−0)2;

whence, squaring, collecting terms and simplifying, (x+a)2 + y2 = (x− a)2+ y2,

4ax=0;

x=0.

We find that the coordinates (x, y) of P are subject to the restriction x=0; this is the equation of the desired locus. The locus is thus seen to be the y-axis, or the mid-perpendicular to AB.

This problem is typical, and the steps taken in solving it should be carefully noted:

To find the locus of a point whose motion is restricted by a given condition:

(1) Choose two convenient perpendicular lines as axes.

(2) Locate every given point by coordinates, to be treated as known constants.

(3) Represent the variable coordinates of any point of the locus by (x, y).

(4) Put the given condition in the form of an equation involving x and y.

(5) Simplify this equation, which is the equation of the required locus.

(6) Identify the locus from its equation.

110. The Equation of the Circle.-To find the equation of a circle with given radius a. First, let the center of the circle be at the origin (Fig. 18), and let P(x, y) be any point of the circle.

The distance OP=√x2+y2 is equal to a, by the definition of the circle.