1.10 Further exercises
Example 12
Determine the equation of the line through each of the following pairs of points. Show that both equations can be written in the form ax + by = c, for some real numbers a, b and c, where a and b are not both zero.
(a) (−2, −4) and (1, 6).
(b) (0, 0) and (7, 3).
Answer
(a) Since (−2, −4) and (1, 6) lie on the line, its gradient is
It follows that the equation of the line is
which can be simplified to
that is,
10x − 3y = −8.
This equation is of the desired form, with a = 10, b = −3 and c = −8. (Any multiple of these numbers is also a valid answer.)

(b) Since the line passes through the origin and the point (7, 3), it has an equation of the form y = mx, for some m. The coordinates of (7, 3) must satisfy the equation y = mx. Thus 3 = 7m, so
.
Hence the equation of the line is
This can be written as
3x − 7y − 0,
which is of the desired form, with a = 3, b = −7 and c = 0.
Example 13
Determine the values of k for which the lines 3x + 4y + 7 = 0 and 2x + ky = 3 are
 (a) parallel,
 (b) perpendicular.
Answer
The gradients of the lines 3 x + 4y + 7 = 0 and 2 x + ky = 3 are and , respectively.
Thus the lines are
(a) parallel if , that is, ;

(b) perpendicular if , that is, .
Example 14
Sketch the lines with the following equations, on a single diagram:
Answer
Example 15
Determine the coordinates of the points of intersection of the lines in Exercise 14.
Answer
Let A be the point of intersection of the lines y = −3x and . We equate the two expressions for y to obtain
Multiplying through by 3 gives
−9x = x + 6,
so
Since A lies on the line y = −3x, it follows that
So the point A has coordinates
Next, let B be the point of intersection of the lines and y − 3 = 3(x − 3). We rewrite the second equation, and equate the two expressions for y to obtain
Multiplying through by 3 and collecting terms gives
so
x = 3.
Since B lies on the line it follows that y = 3. So the point B has coordinates (3, 3).
Finally, let C be the point of intersection of the lines y = −3x and y − 3 = 3(x − 3). We rewrite the second equation, and equate the two expressions for y to obtain
−3x = 3(x − 3) + 3.
Collecting terms gives
6x = 6,
so
x = 1.
Since C lies on the line y = −3x, it follows that y = −3. So the point C has coordinates (1, −3).
Example 16
Find the distances between the vertices of the triangle formed by the points of intersection found in Exercise 15.
Answer
We use the Distance Formula given Section 1.5.
Since , B = (3, 3) and C = (1, −3),
Remark: In the triangle ABC,
AB^{2} + AC^{2} = BC^{2},
so BAC is a right angle.
Example 17
Determine whether each of the pairs of planes given by the following equations intersect, are parallel, or coincide.
(a) x = 1 and y = 2.
(b) z = 1 and z = 3.
Illustrate your answer to each part with a sketch.
Answer
(a) The planes with equations x = 1 and y = 2 are parallel to the (y, z)plane and the (x, z)plane, respectively. They intersect in a line, as shown.
(b) The planes with equations z = 1 and z = 3 are both parallel to the (x, y)plane. They are parallel to each other, as shown.
Example 18
Determine the distance between the points (1, −2, 3) and (−2, 3, −1) in ^{3}.
Answer
We use the Distance Formula given in Section 1.9. The required distance is thus