The straight line y = x–2 rotates about a point where it cuts the x-axis and becomes perpendicular to the straight line ax + by + c = 0. Then its equation is.

(A) ax + by + 2a = 0

(B) ax – by – 2a = 0

(C) bx + ay – 2b = 0

(D) ay – bx + 2b = 0

Slope of the line in the new position is b/a, since it is perpendicular to the line ax + by + c = 0 and it cuts the x-axis. Hence the required line passes through (2, 0) and its slope is b/a.

The required equation is y – 0 = b/a (x-2)

or, ay = bx – 2b or, ay – bx + 2b = 0.

Hence (D) is the correct answer.

Consider the equation
y - y_{1} = m(x-x_{1}). If m and x_{1} are fixed and
different lines are drawn for different values of y_{1}, then.

(A) the line will pass through a fixed point.

(B) there will be a set of parallel lines.

(C)
all
the lines intersect the line x = x_{1}

(D)
all
the lines will be parallel to the line y=y_{1}.

For a fixed value of m, the given
lines form a set of parallel lines all with the slope m,

&⇒
answer (B) is correct, while (A) is incorrect. However this set will not be
parallel to the line y = y_{1}, whose slope is zero, unless

m = 0. And (D) is therefore excluded. If, furthermore, x_{1} is fixed,
the above set of lines will all intersect the line x = x_{1}. Hence
answer (C) is also correct. Hence (B) and (C) are the correct answers.

Let 2x–3y =0 be a given line and P (sinq, 0) and Q (0, cosq) be the two points. Then P and Q lie on the same side of the given line, if q lies in the .

(A) 1st quadrant

(B) 2nd quadrant

(C) 3rd quadrant

(D) 4th quadrant

P and Q lie on the same side if 2sinq and - 3cosq have the same signs i.e. sinq and cosq have opposite signs which is true for the 2nd and 4th quadrant. .

Hence (B) and (D) are the correct answers.

The straight lines of the family x(a + b) + y (a – b) = 2a (a and b being parameters) are

(A) not concurrent

(B) Concurrent at (1, -1)

(C) Concurrent at (1, 1)

(D) None of these

The given equation can be written as

a(x + y – 2) + b(x – y) = 0 or (x + y – 2) + b/a(x – y) = 0

This is a family of lines concurrent at x – y = 0 and x + y – 2 = 0

On solving these two equations we get (1, 1)

Hence (C) is the correct answer.

Drawn from origin are two mutually perpendicular lines forming an isosceles triangle together with the straight line 2x+y = a. Then the area of this triangle is.

(A) _{}

(B) _{}

(C) _{}

(D) None of these

Let the two perpendiculars through the origin intersect 2x+y = a at A and B so that the triangle OAB is isosceles.

OM = length of perpendicular from O
to AB, OM = Hence (C) is the correct answer. . |

A curve with
equation of the form y = ax^{4} + bx^{3}+cx +d has zero
gradient at the point (0, 1) and also touches the x-axis at the
point ( -1, 0). Then the values of x for which the curve has
negative gradients are.

(A) x > -1

(B) x < 1

(C) x < -1

(D) -1 £ x £ 1

y
= ax^{4} + bx^{3} +cx +d .
. . . (1).

y touches x-axis at (-1, 0)

so, ( -1, 0) lies on it and dy/ dx =0

so, 0 = a – b – c +d . . . . (2).

From
(1) dy/ dx = 4ax^{3} + 3bx^{2} +c . . . .
(3).

Hence
_{}

&⇒
- 4a + 3b +c =0 . . . . (4)

Also
_{}= c =0 (since
curve touches (0, 1) )

&⇒ ( 0, 1) also lies on it. Hence d
= 1.

Putting
values of c and d in (2) and (4), solving for a and b we get a = 3, b
= 4. Therefore (3) becomes _{}= 12x^{3}
+12x^{2}

Now dy/dx < 0

&⇒
12 x^{3} + 12x^{2 }< 0

&⇒ 12x^{2} (x+1) < 0

&⇒
x < -1

Hence (C) is the correct answer.

The area of triangle is 5. Two of its vertices are (2, 1) and (3, -2), the third vertex is lying on y = x + 3. The co-ordinates of the third vertex can be.

(a) _{}

(b) _{}

(c) _{}

(d) _{}

As the third vertex lies on the line y = x+3, its co-ordinates are of the form (x, x+3). .

The area of triangle is _{}|4x-4|
= |2x-2|

According to given
condition

&⇒ 2x-2 = ± 5

&⇒
x = -3/2, 7/2

Hence the coordinates of third vertex
can be _{}

Hence (A) and (C) are the correct answers.

The orthocentre of
the triangle formed by the lines 2x^{2} + 3xy – 2y^{2} – 9x +
7y – 5 = 0, 4x + 5y – 3 = 0 lies at

(A) (3/5 , 11/5)

(B) (6/5, 11/5)

(C) (5/6, 11/5)

(D) None of these

The pair of straight lines 2x^{2} + 3xy – 2y^{2}
– 9 x + 7y– 5 = 0 are perpendicular to each other so orthocentre is point of
intersection of these lines. Hence (A) is correct.

If the lines x = a + m, y = -2 and y = mx are concurrent, the least value of |a| is

(A) 0

(B) Ö2

(C) 2Ö2

(D) None of these

Since the lines are concurrent

_{}

&⇒
m^{2} + am + 2 = 0

Since m is real, a^{2}
≥
8, |a| ≥2Ö2

Hence the least value of |a| is 2Ö2

Hence (C) is the correct answer.

If the line y = _{}x
cuts the curve x^{3} + y^{3} +3xy + 5x^{2} + 3y^{2}
+ 4x + 5y – 1 = 0 at the points A, B, C then OA. OB. OC is

(A) _{}

(B) 3_{}+1

(C) _{}

(D) none of these

The Line y=_{}passes
through the origin.

Therefore it can be written as _{}

where r is the distance of any point
(x,y) on y = _{}x from (0, 0)

i.e. _{}will always
lie on the line y = _{}x at a
distance r from (0,0)

Since it cuts the curve

x^{3} + y^{3} + 3xy +
5x^{2} + 3y^{2} + 4x + 5y – 1 = 0

we have _{}

r^{3} _{}

This is a cubic in r which shows that
y = _{}x cuts the
curve at 3 points.

\OA.
OB. OC = _{}

Hence (A) is the correct answer.