Q1.
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
Solution:
(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.
Q2.
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
Solution:
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.
Q3.
Let ax + by + c = 0
be a variable straight line, where a, b and c are 1st, 3rd
and 7th terms of some increasing A.P. Then the variable straight
line always passes through a fixed point which lies on.
(A) x2 +
y2 = 13
(B) x2 +
y2 = 5
(C)
(D) 3x+4y=9
Solution:
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.
Q4.
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
Solution:
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.
Q5.
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
Solution:
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.
Q6.
The area of the
rhombus enclosed by the lines ax ± by ± c = 0 is
(A) 2c2/ab
(B)
2ab/c2
(C)
2c/ab
(D)
None of these
Solution:
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.
Q7.
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
Solution:
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.
Q8.
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
Solution:
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. .
Q9.
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) x2
+ Ax +2A = 0
(B) x2
– 2Ax + 2A = 0
(C) x2
– Ax + 2A = 0
(D)
None of these
Solution:
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
x2 – (a+b)x + ab = 0 or x2
– 2Ax + 2A = 0
Hence (B) is the correct answer.
Q10.
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
Solution:
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),
&⇒
&⇒
a2 b – 2as + 2as
= 0
Since a is real, 4s2
– 8abs ≥
0
&⇒ s ≥ 2ab
Hence the least value
of s = 2ab
Hence (A) is the
correct answer.