Prove that (4, -1), (6, 0), (7,2) and (5, 1) are the vertices of rhombus. Is it a square?.

Let the given points be A, B, C and D respectively. .

Then, coordinates of mid-point of AC are

Coordinates of the mid-points of BD are

Thus, AC and BD have the same mid-point. Hence, ABCD is a parallelogram. .

Now, AB =

BC =

\ AB = BC. .

So, ABCD is a parallelogram whose adjacent sides are equal

Hence, ABCD is a rhombus. .

Now, AC = and BD =.

Clearly, AC > BD. So, ABCD is not a square. .

Let P (sinq, cosq) (0 £ q £ 2p) be a point and let OAB be a triangle with vertices (0, 0),and Find q if P lies inside the DOAB.

Equations of lines
along OA, OB and AB are y = 0, x = 0, and x+y =respectively. Now P and
B will lie on the same side of y = 0 if cosq > 0.
Similarly P and A will lie on the same side of x = 0 if sin q
> 0 and P and O will lie on the same side of x + y = if sin q
+ cosq < .

Hence P will lie inside the D ABC,

if sinq > 0, cosq > 0 and sinq + cosq <.

Now sinq + cosq < i.e. 0 < q + p/4 < p/3 . or < q + p/4 < p Since sinq>0 and cos q > 0, so 0< q< |

A
line intersects the straight lines 5x – y – 4 = 0 and 3x – 4y – 4 = 0 at A and
B respectively. If a point P (1, 5) on the line AB is such that

AP: PB = 2 : 1 (internally), find the point A.

Let AP = 2r. Then PB = r. Let the slope of AB be tanq. . The equation of the line AB is = r, so that A º (1 + 2rcosq, 5 + 2rsinq) and Bº (1 – rcosq, 5 – rsinq) or Aº (1 - 2rcosq, 5 - 2rsinq) and Bº (1 + rcosq, 5 + rsinq) The point ‘A’ lies on the line 5x – y – 4 = 0, So that |

5 (1 + 2rcosq) - (5 +2rsinq) - 4 = 0 ….(i).

The Point ‘B’ lies on the line 3x – 4y – 4 0,

So that 3 (1–rcosq) + 4(5–rsinq) – 4 = 0 ….(ii).

Let rcosq = a and rsinq = b. .

&⇒
10a – 2b – 4 = 0 and –3a
– 4b + 19 = 0

On solving for a and b, we get

Hence the point A is

Similarly when A º (1 - 2rcosq, 5 - 2rsinq)

and B º (1 + rcosq, 5 + rsinq), we get

- 10a + 2b - 4 = 0 and 3a +4b +19 = 0

&⇒
a
= , b
=

Hence the point A is .

Four points A(a, 0), B(b, 0), C(g, 0) and D(d, 0), with a < b < g
< d or

a > b > g > d
, are such that a and b are the roots of ax^{2}
+2hx + b = 0 and g and d are the roots of a¢x^{2} +2h¢x + b¢ = 0. If = l, = m and

ab¢ + ba¢ = 2hh¢, show that l
+ m =0.

We have a + b = -2h/a, ab = b/a. Also, g + d = , gd =

&⇒ l
= . Also, m
= , l
+ m = +

= =

= =

= = 0.

Find
the points farthest and nearest on the curve 5x^{2} + 5y^{2} +
6xy– 8 = 0 from origin. Also determine the equations of the lines through the
origin on which the points are occurring. .

Any line through the
origin is Any point on this line
is

(r cosq, r sinq). If this point lies on the given
curve,

then 5r^{2} + 6r^{2}sinq
cosq = 8

&⇒
r^{2 }=

&⇒
r_{max} = 2 (for q= ) and r_{min} =
1 (for q = p/4).

The required points are
() and (). The lines are

y = – x and y = x.

&⇒