The equation of the diagonal, through the origin, of the quadrilateral formed by the lines x = 0, y = 0, x + y = 1 and 6x + y = 3 is given by

(A) 3x –y = 0

(B) 3x – 2y = 0

(C) x – y = 0

(D) 3x – 4y = 0

(0, 0) is the one end of the diagonal. The intersection of the other two lines, x + y = 1 and 6x + y = 3 is (2/5, 3/5) which is the other end of the diagonal. The equation of the line that passes through these two points is 3x - 2y = 0.

Hence (B) is the correct answer.

If a, b, c are in A.P, then ax + by + c = 0 represents.

(A) a single line

(B) (B) a family of concurrent lines

(C) a family of parallel lines

(D) (D) none of these

Since a, b, c are in A.P. .

&⇒
2b = a + c

&⇒ a – 2b + c = 0

&⇒
Family of lines is concurrent at the point (1, -2)

Hence (B) is the correct answer.

Let ax + by + c = 0
be a variable straight line, where a, b and c are 1^{st}, 3^{rd}
and 7^{th} terms of some increasing A.P. Then the variable straight
line always passes through a fixed point which lies on.

(A) x^{2 }+
y^{2} = 13

(B) x^{2 }+
y^{2} = 5

(C) _{}

(D) 3x+4y=9

Let A.P. be l,l + m, l+2m, l+3m,….

Given that l = a, l+2m = b, l+6m = c

Clearly 2a-3b+c=0

So fixed point is (2,-3)

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

Let ABC be a triangle with equations of the sides AB, BC and CA respectively x – 2 =0, y – 5 = 0 and 5x + 2y –10 = 0. Then the orthocentre of the triangle lies on the line.

(A) x–y = 0

(B) 3x –y =1

(C) 4x +y = 13

(D) x–2y = 1

The given triangle is a right angled triangle. Hence the orthocentre is the vertex containing the right angle. .

&⇒
orthocentre is (2, 5) which lies on the lines 3x – y = 1,

and 4x + y = 13.

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

Two
lines are given by (x – 2y)^{2} + k (x – 2y) = 0. The value of k, so
that the distance between them is 3, is .

(A) k = 0

(B) k = ± 3Ö5

(C) k = -5

(D) k = 3

The lines x – 2y = 0
and x – 2y + k = 0 are parallel. The distance between these two lines = _{}

&⇒
k = ± 3Ö5

Hence (B) is the correct answer.

The area of the rhombus enclosed by the lines ax ± by ± c = 0 is

(A) 2c^{2}/ab

(B)
2ab/c^{2}

(C) 2c/ab

(D) None of these

The four sides of the rhombus are

ax + by + c = 0, ax + by –c = 0, ax –by + c = 0, ax – by – c = 0 .

On solving these equations, we get the vertices as A (c/a, 0),

B (0, c/b), C(-c/a, 0) and D(0, -c/b).

The length of the
diagonal AC is 2c/a and that of the diagonal BD

is 2c/b.

Therefore the area of
the rhombus is 1/2 _{}

Hence (A) is the correct answer.

The medians AD and BE of a triangle ABC with vertices A(0, b), B(0, 0) and C(a, 0) are perpendicular to each other if

(A) b = Ö2a

(B) a = Ö2b

(C) b = -Ö2a

(D) a = -Ö2b

The coordinates of the mid-point D of
BC are _{}and the coordinates
of the mid-point E of CA are _{}. If AD and BE
are mutually perpendicular, then slope of AD ´ slope of BE =
-1

&⇒_{}

&⇒
a = ± Ö2b. Hence (B), (D) are correct answer.

If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is

(A) square

(B) a circle

(C) straight line

(D) two intersecting lines

Let the two perpendicular lines be the coordinate axes and let the point be P(h, k) . Then sum of the distances of P(h, k) from the coordinate axes is |h| +|k|. It is given that |h| +|k| = 1. .

So locus of (h,k) is |x| +|y|=1. .

This gives four lines x ± y = 1, -x +y = 1, -x - y = 1,

which enclose a square. .

Hence (A) is the correct answer. .

The quadratic equation whose roots are the x and y intercepts of the line passing through (1, 1) and making a triangle of area A with the coordinate axes is

(A) x^{2}
+ Ax +2A = 0

(B) x^{2}
– 2Ax + 2A = 0

(C) x^{2}
– Ax + 2A = 0

(D) None of these

Equation of the line having intercepts a and b on x and y axes, respectively, is x/a + y/b = 1 . . . . . (1).

It passes through (1, 1)

&⇒
1/a + 1/b = 1

Since the area of the triangle formed by the lines and the axes is A,

ab = 2A . . . . .(2).

From (1) & (2) we get a + b = 2A

Here a, b are the roots of

x^{2} – (a+b)x + ab = 0 or x^{2}
– 2Ax + 2A = 0

Hence (B) is the correct answer.

Through
the point P(a, b), where ab > 0, the straight line _{}is drawn so as
to form with coordinate axes a triangle of area s. If ab > 0, then least
value of s is

(A) 2ab

(B) 1/2 ab

(C) ab

(D) none of these

Given Pº (a, b)

Given line is _{} .
. . . . (1)

If line (1) cuts x and y axes at A and B respectively, then Aº (a, 0) and B º (0, b) . Also the area of DOAB = s.

(1/2)ab = s

&⇒
ab = 2s

Since line (1) passes through P(a, b),

_{}

&⇒
_{}

&⇒
a^{2} b – 2as + 2as
= 0

Since a is real, 4s^{2}
– 8abs ≥
0

&⇒ s ≥ 2ab

Hence the least value of s = 2ab

Hence (A) is the correct answer.