Example 1
If
the coordinates of the mid-points of the sides of a triangle are (1, 2),
(0,–1) and (2, -1). Find the coordinates of its vertices. .
Solution:
Let A(x1, y1),
B(x2, y2) and C(x3, y3) be the
vertices of DABC. Let
D(1, 2), E(0, –1) and F(2, -1) be the mid-points of sides BC, CA and AB
respectively. .
Since D is the mid-point of BC
\
=1 and 
= 2
&⇒
x2 + x3 =2 and y2 + y3 =4 .
. . (i).
Similarly, E and F are the mid-points
of CA and AB respectively,
we have
x1
+ x3 = 0 and y1 + y3 = –2 .
. . (ii).
x1
+ x2 =4 and y1 + y2 = -2 .
. . (iii).
From
(i), (ii) and (iii), we get
x1
+ x2 + x3 =3 and y1 + y2 + y3
= 0 . . . (iv).
From
(i) and (iv), we get x1 + 2 = 3 and y1 + 4 = 0
\ x1 = 1 and y1
= -4, so, the coordinates of A are (1, -4)
From
(ii) and (iv), we get x2 + 0 = 3 and y2 –2 = 0
\ x2 =3 and y2
= 2, so, coordinates of B are (3, 2)
From
(iii) and (iv), we get x3 + 4 = 3 and y3 –2 = 0
\ x3 = -1 and y3
= 2, so, coordinates of C are (-1, 2)
Hence,
the vertices of the triangle ABC are
A(1,
-4), B(3, 2) and C(-1, 2). .
Example 2
Let the opposite vertices of a square
be ( 3, 4) and ( 1, -1). Find the coordinates of the remaining vertices. .
Solution:
|
Let ABCD be a square and let
A(3, 4) and C(1, -1) be the given vertices. Let B(x, y) be the
unknown vertex. .
Then, AB = BC
&⇒
AB2 = BC2
&⇒
(x – 3)2 + (y – 4)2 = (x –1)2 + (y +1)2
&⇒
4x + 10 y – 23 = 0
&⇒
x =  . . .
(i)
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In triangle ABC, we have AB2
+BC2 = AC2
(x –3)2 + (y –4)2
+ (x –1)2 +(y +1)2 = (3 –1)2 +(4 +1)2
&⇒
x2 +y2 - 4x - 3y –1 = 0 . . . (ii) .
Substituting the value of x from (i)
into (ii), we get
+ y2 –
– 3y – 1 = 0
&⇒
4y2 -12y + 5 = 0
&⇒
(2y –1) (2y –5) = 0
&⇒ y = 
,
hence x = 9/2 or x = -1/2
respectively. .
The required points are (9/2, 1/2)
and (-1/2, 5/2). .
Example 3
Find the equation of the straight lines
passing through (-2, -7) and having intercept of length 3 units between the
straight lines 4x + 3y = 12 and
4x + 3y = 3.
Solution:
|
Distance between the
two given parallel lines
=  =  = 
tanq = 
&⇒
tanq = 
|
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Slope of the parallel
lines = 
= m2
Also tanq
= ±
&⇒ 
&⇒
m1 + 
.
The slopes are
(1) m1 =
-7/24
(2) m1 = ¥(the
line is parallel to the y - axis)
The required equations of the lines
are 7x + 24y + 182 = 0 and x+2 =0.
Example 4
The equations of the perpendicular
bisectors of the sides AB and AC of a triangle ABC are respectively x – y + 5 =
0 and x + 2y = 0. If the
co-ordinates of A are (1, –2), find the equation of BC.
Solution:
|
Let OE and OF be the
perpendicular bisectors of AB and AC respectively. .
Let B be (x1,
y1) and C be (x2, y2). E is the mid-point
of AB and F is the mid-point of AC. .
E º
 , F º
Since E and F lie on
OE and OF respectively, x1 - y1 + 13 = 0 . . . (1).
and
x2 + 2y2 - 3 = 0 . . . (2).
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Also, slope of AB = -1
and that of AC = 2, so that
x1 + y1 + 1 = 0 .
. . (3).
and 2x2 - y2 -
4 = 0 . . . (4).
Solving these equations, we get the
co-ordinates of B and C as
B º (-7, 6) and C º
(11/5, 2/5)
&⇒
Equation of BC is 14x + 23y - 40 = 0.
Example 5
Find a, if (a, a2) lies inside the
triangle having sides along the lines
2x + 3y = 1, x + 2y – 3 = 0, 6y = 5x – 1.
Solution:
The vertices of the triangle ABC
formed by the given lines are
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A º
(-7, 5), B º (5/4, 7/8)
C º
(1/3, 1/9)
Sign of A w.r.t.
BC is –ve.
If P lies
in-side the DABC, then sign of P will be same as
sign of A w.r.t the line BC .
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&⇒
5a
- 6a2
- 1 < 0 . . . . . (1).
Similarly 2a
+ 3a2
- 1 > 0 . . . . . (2).
and, a + 2a2 - 3 < 0 .
. . . . (3).
Solving, (1), (2) and (3) for a
and then taking intersection
we get a ∈
(1/2, 1) È (-3/2, -1) .
