COORDINATE GEOMETRY
Q. Find the ratio in which the line segment joining the points (–3, 10) and
(6, –8) is divided by (–1 ,6).
Sol.
A
P
B
(–3, 10)
(–1, 6)
(6, –8)
A (–3, 10),
B (6, –8), P(–1, 6)
=
m1
(6)
+
m2
(–3)
x =
m1
x2
+
m2
x1
m1 + m2
By section formula for internal division,
–1
∴
∴
=
6m1
–
3m2
∴
=
6m1
–
3m2
–
m2
=
7m1
2m2
2
7
=
m1
m2
m1 : m2
2 : 7
=
∴
∴
∴
m1 + m2
y1 = 10
x1 = –3,
y = 6
x = –1,
y2 = –8
x2 = 6,
To find : m1 : m2
–(m1 + m2)
–m1
∴
=
6m1
+
m1
+
3m2
–m2
=
m1
m2
i.e.
2
7
Let, P (–1,6) divides seg AB internally in the ratio m1 : m2.
Let the co-ordinates of A be (x1, y1)
Let the co-ordinates of B be (x2, y2)
Let the co-ordinates of P be (x, y)
Which formula should we apply here ?
Section formula for Internal Division.
+
m1y2
m2y1
+
m2
m1
y
=
,
+
m1x2
m2x1
+
m2
m1
x
=
To find - m1 : m2
P lies on X-axis,
∴ Its Y-coordinate is 0.
∴ P(x, 0)
A(1, –5)
Xl
B(–4, 5)
X
Q. Find the ratio in which the line segment joining A (1, –5) and B (–4, 5) is
divided by the X–axis. Also find the coordinates of the point of division.
Sol.
By using section Formula,
∴
A (1, –5),
B (–4, 5) P(x, 0)
(x, 0)
=
m1
(5)
+
m2
(–5)
y =
m1
y2
+
m2
y1
m1 + m2
0
∴
∴
=
5m1
–
5m2
=
5m2
5m1
m1 : m2
1 : 1
=
∴
∴
∴
m1 + m2
x1 = 1,
x2 = –4,
0
=
m1
m2
∴
1
1
Let, P (x,0) divides seg AB internally in the ratio m1 : m2.
Let the co-ordinates of A be (x1, y1)
Let the co-ordinates of B be (x2, y2)
Let the co-ordinates of P be (x, y)
Which formula should we apply here ?
Section formula for Internal Division.
+
m1y2
m2y1
+
m2
m1
y
=
,
+
m1x2
m2x1
+
m2
m1
x
=
=
m2
m1
x = x ,
y = 0
y2 = 5
y1 = –5
m1 : m2
P
Q. Find the ratio in which the line segment joining A (1, –5) and B (–4, 5) is divided by the X–axis. Also find the coordinates of the point of division.
Sol.
∴
X– axis divides AB in the ratio 1 : 1
A(1, –5)
Xl
B(–4, 5)
X
P(x, 0)
m1 : m2
1 : 1
=
A (1, –5),
B (–4, 5) P(x, 0)
y1 = –5
x1 = 1,
y = 0
x = x ,
y2 = 5
x2 = –4,
=
1
(–4)
+
1
(1)
x =
m1
x2
+
m2
x1
m1 + m2
x
∴
1 + 1
=
–4
+
1
x
∴
2
=
–3
x
∴
2
m1 : m2
= –1.5
∴ P (–1.5, 0)