COORDINATE GEOMETRY
points P (2, –3) and Q (10, y) is 10 units.
Soln.
PQ
=
10
Now
∴
(10
–
2)
2
+
[y
2
–
(–3)
]
10
=
∴
8
2
+
(y
2
+
3)
10
=
∴
100
=
64
+
(y
+
3)
2
∴
100
=
y
+
6y
+
9
2
–
64
∴
y
+
6y
+
9
2
–
36
=
0
∴
y
+
6y
–
2
27
=
0
∴
y
+
9y
–
3y
2
–
27
=
0
∴
y
(y
+
9)
–
3
(y
+
9)
=
0
∴
(y
+
9)
(y
–
3)
=
0
∴
y
=
–9
or
y
=
3
P(2, –3), Q(10, y)
x1 = 2,
y1 = –3
x2 = 10,
y2 = y
[Squaring throughout]
Let the coordinates of Q be (x2, y2).
Let the coordinates of P be (x1, y1).
+
–
–
(
)
(
)
x2
x1
y2
y1
2
2
PQ
=
What is the formula to find distance between two points ?
+
–
–
(
)
(
)
x2
x1
y2
y1
2
2
P
(x,y)
A
(2,–5)
B
(–2,9)
Q. Find the point on the x– axis which is equidistant from
(2, –5) and (–2, 9).
X
–
(x
2)2
=
+
By distance formula,
–
[0
(–5)]2
–
[x
(–2)]2
+
–
(0
9)2
Sol. Let, A (2, –5) B (–2, 9)
(x,0)
AP
=
BP
AP = BP
∴
x2 = x,
y2 = 0
x1 = 2 ,
y1 = –5
x1 = –2,
y1 = 9
x2 = x,
y2 = 0
∴
(x
–
2)
2
+
(5)
2
=
(x
+
2)
2
+
(–9)
2
∴
x
–
4x
+
4
2
+
25
=
x
+
4x
+
4
2
+
81
∴
– 4x
–
4x
=
85
–
29
∴
–8x
=
56
∴
The coordinates of P are (–7,0) .
∴
– 4x
+
29
=
4x
+
85
=
x
∴
56
–8
=
–7
7
Squaring on both sides
Let the coordinates of A be (x1, y1)
Let the coordinates of P be (x2, y2)
Let the coordinates of B be (x1, y1)
Let the coordinates of P be (x2, y2)
Which is the formula to find length of AP and BP?
+
–
–
(
)
(
)
x2
x1
y2
y1
2
2
What do we know about any point on x-axis ?
Its y co-ordinate is zero.
Let P be the point on x-axis
Let A (2,–5) and B (–2,9)
Here, point P is on x-axis
So, y = 0
Let the co-ordinates of P be (x,y)
We need to find
co-ordinates of point P.
EQUIDISTANT means ‘Equal Distance’.