COORDINATE GEOMETRY
Sol.
AP
AB
∴
3AP
=
AP
A
P
B
(2, 1)
(x, y)
(5, –8)
=
∴
AP
AP
+
BP
1
3
=
1
3
+
BP
∴
3AP
–
AP
=
BP
∴
2AP
=
BP
AP
BP
=
1
2
∴
= 3
y =
m1
y2
+
m2
y1
m1 + m2
x =
m1
x2
+
m2
x1
m1 + m2
A (2, 1),
B (5, –8)
= 1 : 2
=
1
(5)
+
2
(2)
1
+
2
=
5
+
4
3
9
3
=
1
(–8)
+
2
(1)
1
+
2
=
–8
+
2
3
–6
3
By using section formula, we get
=
y
Sol.
m1:m2
∴
=
x
∴
Let the co-ordinates of B be (x2, y2)
x1 = 2,
y1 = 1
x2 = 5,
y2 = –8
P =
3
,
– 2
Let the co-ordinates of A be (x1, y1)
Which formula is used to find co-ordinates of P?
,
+
m1x2
m2 x1
+
m2
m1
x
=
+
m1y2
m2y1
+
m2
m1
y
=
Section formula for Internal Division.
A
P
B
(2, 1)
(x, y)
(5, –8)
= – 2
Sol.
A
P
B
(2, 1)
(3, –2)
(5, –8)
P lies on the line 2x – y + k = 0
∴
Coordinates of point P satisfy the given
equation 2x – y + k = 0
∴
We substitute x = 3, y = – 2
2
(3)
–
(– 2)
+
k
=
0
∴
6
+
2
+
k
=
0
∴
8
+
k
=
0
∴
k
=
– 8