 Vectors and conics

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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).

• (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.

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:  Example 15

Determine the coordinates of the points of intersection of the lines in Exercise 14.

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.

We use the Distance Formula given Section 1.5.

Since , B = (3, 3) and C = (1, −3), Remark: In the triangle ABC,

AB2 + AC2 = BC2,

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.

• (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.

We use the Distance Formula given in Section 1.9. The required distance is thus M208_1