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. .

Let A(x_{1}, y_{1}),
B(x_{2}, y_{2}) and C(x_{3}, y_{3}) 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

&⇒
x_{2} + x_{3} =2 and y_{2} + y_{3} =4 .
. . (i).

Similarly, E and F are the mid-points of CA and AB respectively,

we have

x_{1}
+ x_{3} = 0 and y_{1} + y_{3} = –2 .
. . (ii).

x_{1}
+ x_{2} =4 and y_{1} + y_{2} = -2 .
. . (iii).

From (i), (ii) and (iii), we get

x_{1}
+ x_{2} + x_{3} =3 and y_{1} + y_{2} + y_{3}
= 0 . . . (iv).

From
(i) and (iv), we get x_{1} + 2 = 3 and y_{1} + 4 = 0

\ x_{1} = 1 and y_{1}
= -4, so, the coordinates of A are (1, -4)

From
(ii) and (iv), we get x_{2} + 0 = 3 and y_{2} –2 = 0

\ x_{2} =3 and y_{2}
= 2, so, coordinates of B are (3, 2)

From
(iii) and (iv), we get x_{3} + 4 = 3 and y_{3} –2 = 0

\ x_{3} = -1 and y_{3}
= 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). .

Let the opposite vertices of a square be ( 3, 4) and ( 1, -1). Find the coordinates of the remaining vertices. .

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 |

In triangle ABC, we have AB^{2}
+BC^{2} = AC^{2}

(x –3)^{2} + (y –4)^{2}
+ (x –1)^{2} +(y +1)^{2} = (3 –1)^{2} +(4 +1)^{2}

&⇒
x^{2} +y^{2} - 4x - 3y –1 = 0 . . . (ii) .

Substituting the value of x from (i) into (ii), we get

+ y^{2 }–– 3y – 1 = 0

&⇒
4y^{2} -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). .

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.

Distance between the two given parallel lines = = = tanq = |

Slope of the parallel
lines = = m_{2}

Also tanq
= ±

&⇒

&⇒
m_{1} + .

The slopes are

(1) m_{1} =
-7/24

(2) m_{1} = ¥(the
line is parallel to the y - axis)

The required equations of the lines are 7x + 24y + 182 = 0 and x+2 =0.

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.

Let OE and OF be the perpendicular bisectors of AB and AC respectively. . Let B be (x E º , F º Since E and F lie on
OE and OF respectively, x and
x |

Also, slope of AB = -1 and that of AC = 2, so that

x_{1} + y_{1} + 1 = 0 .
. . (3).

and 2x_{2} - y_{2} -
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.

Find a, if (a, a^{2}) lies inside the
triangle having sides along the lines

2x + 3y = 1, x + 2y – 3 = 0, 6y = 5x – 1.

The vertices of the triangle ABC formed by the given lines are

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 . |

&⇒
5a
- 6a^{2}
- 1 < 0 . . . . . (1).

Similarly 2a
+ 3a^{2}
- 1 > 0 . . . . . (2).

and, a + 2a^{2} - 3 < 0 .
. . . . (3).

Solving, (1), (2) and (3) for a and then taking intersection

we get a ∈ (1/2, 1) È (-3/2, -1) .