COORDINATE GEOMETRY
Introduction
The distance of a point from the y-axis is called its x-coordinate, or abscissa.
The distance of a point from the x-axis is called its y-coordinate, or ordinate.
The coordinates of a point on the x-axis are of the form (x, 0), and of a point on the y-axis are of the form (0, y).
So, the distance between these two points can be found out by using Pythagoras Theorem
i.e distance=$√(x^2+y^2)$
distance= $x_2-x_1$
distance= $y_2-y_1$
distance(OP)= $√(x^2+y^2)$
distance= $√{(x_2-x_1)^2+(y_2-y_1)^2}$
Distance between any two points $P(x_1, y_1)$ and $Q(x_2, y_2)$.
Draw PR and QS perpendicular to the x-axis. A perpendicular from the point Q on PR is drawn to meet it at the point T (see Fig.).
Then, OR = $x_1$, OS = $x_2$. So, RS =$x_1-x_2$ = QT.
Also, SQ = $y_2$ =RT , PR = $y_1$. So, PT =$y_1-y_2$
Now, applying the Pythagoras theorem in ∆ PTQ, we get
$PQ^2 = PT^2 + QT^2$
${(x_1-x_2)^2+(y_1-y_2)^2}$
Therefore, PQ=$√{(x_1-x_2)^2+(y_1-y_2)^2}$
which is called the
Example:1 Consider the points P(6, 4) and Q(–5, –3) . Find the distance between PQ. Since distance is given as $√{(x_2-x_1)^2+(y_2-y_1)^2}$ Hence, $√{(6-(-5))^2+(4-(-3))^2}$ $√{(11)^2+(7)^2}$ $√{170}$
Example: Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed.Solution : Let us apply the distance formula to find the distances PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2, 3) are the given points. We : PQ =$√((3 + 2)^2 + (2 + 3)^2)$ = $√(5^2+5^2)$ =$√50$ =7.07 QR =$√((-2-2)^2 + (-3 -3)^2)$ = $√{(-4)^2+(-6)^2}$ =$√52$ =7.21 PQ =$√((3 - 2)^2 + (2 - 3)^2)$ = $√{1^2+(-1)^2}$ =$√2$ =1.41 Since the sum of any two of these distances is greater than the third distance, therefore, the points P, Q and R form a triangle. Also, $PQ^2 + PR^2 = QR^2$, by the converse of Pythagoras theorem, we have ∠ P = 90°. Therefore, PQR is a right triangle.Example: Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.Solution : Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points. One way of showing that ABCD is a square is to use the property that all its sides should be equal and both its digonals should also be equal. Now, AB=$√{(1-4)^2+(7-2)^2}$=$√{(-3)^2+(5)^2}$=$√{9+25}$=$√{34}$ BC=$√{(4+1)^2+(2+1)^2}$=$√{(5)^2+(3)^2}$=$√{25+9}$=$√{34}$ CD=$√{(-1 + 4)^2+(-1 -4)^2}$=$√{(3)^2+(-5)^2}$=$√{9+25}$=$√{34}$ DA=$√{(1 + 4)^2+(7 -4)^2}$=$√{(5)^2+(3)^2}$=$√{25+9}$=$√{34}$ AC=$√{(1 + 1)^2+(7 +1)^2}$=$√{(2)^2+(8)^2}$=$√{4+64}$=$√{68}$ BD=$√{(4 + 4)^2+(2 -4)^2}$=$√{(8)^2+(-2)^2}$=$√{64+4}$=$√{68}$ Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Thereore, ABCD is a square.
Example : Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).Solution : Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5). We are given that AP = BP. So, AP2 = BP2 i.e., $(x – 7)^2 + (y – 1)^2 = (x – 3)^2 + (y – 5)^2$ i.e., $x^2 – 14x + 49 + y^2 – 2y + 1 = x^2 – 6x + 9 + y^2 – 10y + 25$ i.e., $x – y = 2$ which is the required relation.
Example : Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).Solution : We know that a point on the y-axis is of the form (0, y). So, let the point P(0, y) be equidistant from A and B Then, $(6 – 0)^2 + (5 – y)^2 = (– 4 – 0)^2 + (3 – y)^2$ i.e., $36 + 25 + y^2 – 10y = 16 + 9 + y^2 – 6y$ i.e., $4y = 36$ i.e., $y = 9 $ So, the required point is (0, 9). Let us check our solution : AP =$√{(6-0)^2+(5-9)^2}$ =$√{36+16}$=$√52$ BP =$√{(-4-0)^2+(3-9)^2}$ =$√{16+36}$=$√52$ Note : Using the remark above, we see that (0, 9) is the intersection of the y-axis and the perpendicular bisector of AB.
CONCEPT