Find the coordinates of a point whose distance from (1, 0) is V10 and whose distance
from (5, 4) is V10.
Answers
Step-by-step explanation:
It is expected that students have met plotting points on the plane and have plotted points from tables of values of both linear and non linear functions.
The number plane (Cartesian plane) is divided into four quadrants by two perpendicular axes called the x-axis (horizontal line) and the y-axis (vertical line). These axes intersect at a point called the origin. The position of any point in the plane can be represented by an ordered pair of numbers (x, y). These ordered pairs are called the coordinates of the point.
The point with coordinates (4, 2) has been plotted on the Cartesian plane shown. The coordinates of the origin are (0, 0).
Once the coordinates of two points are known the distance between the two points and midpoint of the interval joining the points can be found.
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DISTANCE BETWEEN TWO POINTS
Distances are always positive, or zero if the points coincide. The distance from A to B is the same as the distance from B to A. We first find the distance between two points that are either vertically or horizontally aligned.
EXAMPLE
Find the distance between the following pairs of points.
a A(1, 2) and B(4, 2)b A(1, −2) and B(1, 3)
SOLUTION
a The distance AB = 4 − 1 = 3
Note: The distance AB is obtained from
the difference of the x-coordinates of
the two points.
b The distance AB = 3 − (−2) = 5
Note: The distance AB is obtained from
the difference of the y-coordinates of the
two points.
The example above considered the special cases when the line interval AB is either horizontal or vertical. Pythagoras’ theorem is used to calculate the distance between two points when the line interval between them is neither vertical nor horizontal.
The distance between the points A(1, 2) and B(4, 6) is calculated below.
AC = 4 − 1 = 3 and BC = 6 − 2 = 4.
By Pythagoras’ theorem,
AB2 = 32 + 42 = 25
And so AB = 5
The general case
We can obtain a formula for the length of any interval. Suppose that P(x1, y1) and Q(x2, y2) are two points.
Form the right-angled triangle PQX, where X is the point (x2, y1),
PX = x2 − x1 or x1 − x2 and QX = y2 − y1 or y1 − y2
depending on the positions of P and Q.
By Pythagoras’ theorem:
PQ2 = PX2 + QX2
= (x2 − x1)2 + (y2 − y1)2
Therefore PQ = QP =
Note that (x2 − x1)2 is the same as (x1 − x1)2 and therefore it doesn’t matter whether we go from P to Q or from Q to P − the result is the same.
EXAMPLE
Find the distance between the points A(−4, −3) and B(5, 7).
SOLUTION
In this case, x1 = −4, x2 = 5, y1 = −3 and y2 = 7.
AB2 = (x2 − x1)2 + (y2 − y1)2
= (5 − (−4))2 + (7 − (−3))2
= 92 + 102
= 181
Thus, AB =
Note that we could have chosen x1 = 5, x2 = −4, y1 = 7 and y2 = −3 and still obtained the same result. As long as (x1, y1) refers to one point and (x2, y2) the other point, it does not matter which one is which.
EXERCISE 1
Show that the distance between the points A(a, b) and B(c, d) is the same as the
distance between
the points P(a, d) and Q(c, b)
the points U(b, a) and V(d, c)
Illustrate both of these.
EXERCISE 2
The distance between the points (1, a) and (4, 8) is 5. Find the possible values of a and use a diagram to illustrate.
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THE MIDPOINT OF AN INTERVAL
The coordinates of the midpoint of a line interval can be found using averages as we will see.
We first deal with the situation where the points are horizontally or vertically aligned.
EXAMPLE
Find the coordinates of the midpoint of the line interval AB, given:
aA(1, 2) and B(7, 2) b A(1, −2) and B(1, 3)
SOLUTION
a AB is a horizontal line interval, the
midpoint is at (4, 2), since 4 is halfway
between 1 and 7.
Note: 4 is the average of 1 and 7, that is, 4 = .
b The midpoint of AB has coordinates 1, G4t2.pdf.
Note that is the average of 3 and −2.
When the interval is not parallel to one of the axes we take the average of the x-coordinate and the y-coordinate. This is proved below.
Let M be the midpoint of the line AB. Triangles AMS and MBT are congruent triangles (AAS), and so AS = MT and MS = BT.
Hence the x-coordinate of M is the average of 1 and 5.
x = = 3
The y coordinate of M is the average of 2 and 8.
x = = 5
Thus the coordinates of the midpoint M are (3, 5).
The general case
We can find a formula for the midpoint of any interval. Suppose that P(x1, y1) and Q(x2, y2)are two points and let M(x, y) be the midpoint.
Triangles PMS and MQT are congruent triangles (AAS), and so PS = MT and MS = QT.
Hence the x-coordinate of M is the average of x1 and x2, and y-coordinate of M is the average of y1 and y2. Therefore
x = G4t6.pdf and y =
Midpoint of an interval
The midpoint of an interval with endpoints P(x1, y1) and Q(x2, y2) is G4t8.pdf,.
Take the average of the x-coordinates and the average of the y-coordinates.
EXAMPLE
Find the coordinates of the midpoint of the line interval joining the points (6, 8) and (−3, 2).
SOLUTION
The midpoint has coordinates, G4t10.pdf, G4t11.pdf = , 5