Mathematics - Cartesian Coordinate System
Exam Duration: 45 Mins Total Questions : 30
The equation of the line passing through (1, 2) and perpendicular to X + Y+ 7 = 0 is
- (a)
y - x + 1 = 0
- (b)
y - x - 1 = 0
- (c)
y - x + 2 = 0
- (d)
y - x - 2 = 0
The equation of the line perpendicular to the line X - 7Y + 5 = 0 and having X-intercept 3, is
- (a)
7x + y - 21 = 0
- (b)
6x + y - 19 = 0
- (c)
5x + 2y - 21 = 0
- (d)
6x + 7y - 25 = 0
If k is a parameter, then the equation of the family of lines parallel to the line 3X + 4Y + 5 = 0 is
- (a)
4x - 3y + k = 0
- (b)
3x - 4y + k = 0
- (c)
3x + 4y + k = 0
- (d)
4x + 3y + k = 0
If lines aX + bY + c = 0, where 3a + 2b + 4c = 0 and \(a,b,c\varepsilon R\), then the given set of lines are concurrent at the point
- (a)
(3, 2)
- (b)
(2, 4)
- (c)
(3, 4)
- (d)
(3/4, 1/2)
The distance between the lines 3X + 4Y = 9 and 6X + 8Y = 15 is
- (a)
\(\frac { 3 }{ 10 } \)
- (b)
\(\frac { 2 }{ 9 } \)
- (c)
\(\frac { 1 }{ 4} \)
- (d)
\(\frac { 1 }{ 3} \)
If p is the length of perpendicular from origin to the line whose intercept on the axes are a and b, then \(\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } \) is equal to
- (a)
\(\frac { 1 }{ { p }^{ 3 } } \)
- (b)
\(\frac { 1 }{ { p }}\)
- (c)
\(\frac { 1 }{ { p }^{ 2 } } \)
- (d)
\(p\)
If the equation of the sides of a triangle are X + Y = 2, Y = X and \(\sqrt{3}\) Y + X = 0, then which of the following is an exterior point of the triangle?
- (a)
Orthocentre
- (b)
Incentre
- (c)
Centroid
- (d)
None of the above
If p is the length of perpendicular from the origin on the line \(\frac { X }{ a } +\frac { Y }{ b } =1\) and a2 ,p2 and b2 are in AP, then a4+b4 is equal to
- (a)
1
- (b)
2
- (c)
3
- (d)
0
If one of the diagonal of a square is along the line X = 2Y and one of its vertices is (3, 0), then its side through this vertex nearer to the origin is given by the equation
- (a)
y - 3x + 9 = 0
- (b)
3y + x - 3 = 0
- (c)
x - 3y - 3 = 0
- (d)
3x + y - 9 = 0
All points lying inside the triangle formed by the points (1, 3), (5, 0) and (- 1, 2) satisfy
- (a)
\(3x+2y\ge 0\)
- (b)
\(2x+y-13\ge 0\)
- (c)
\(2x-3y-12\ge 0\)
- (d)
\(-2x+y\ge 0\)
If equations \(\left( b-c \right) X+\left( c-a \right) Y+\left( a-b \right) =0\) and \(\left( { b }^{ 3 }-{ c }^{ 3 } \right) X+\left( { c }^{ 3 }-{ a }^{ 3 } \right) Y+\left( { a }^{ 3 }-{ b }^{ 3 } \right) =0\) represent same line, then
- (a)
a = b = c
- (b)
b = c
- (c)
c = a
- (d)
\(a+b+c\neq 0\)
If \(\frac { 2 }{ 1!9! } +\frac { 2 }{ 3!7! } +\frac { 1 }{ 5!5! } =\frac { { 2 }^{ m } }{ n! } ,\) then orthocentre of the triangle having sides X - Y + 1 = 0, X + Y + 3 = 0 and 2X + 5Y - 2 = 0 is
- (a)
(2m - 2n, m - n)
- (b)
(2m - 2n, n - m)
- (c)
(2m - n, m + n)
- (d)
(2m - n, m - n)
Let a,b,c and d be non-zero numbers. If the point of intersection of the lines 4aX + 2aY + c = 0 and 5bX + 2bY + d = 0 lie in the fourth quadrant and is equidistant from the two axes, then
- (a)
3bc - 2ad = 0
- (b)
3bc + 2ad = 0
- (c)
2bc - 3ad = 0
- (d)
2bc + 3ad = 0
If the line 2X + Y = k passes through the point which divides the line segment joining the points (1, 1) and (2, 4) in the ratio 3 : 2, then k is equal to
- (a)
\(\frac{29}{5}\)
- (b)
5
- (c)
6
- (d)
\(\frac{11}{5}\)
The lines p \(\left( { p }^{ 2 }+1 \right) X-Y+q=0\) and \(\left( { p }^{ 2 }+1 \right) ^{ 2 }X+\left( { p }^{ 2 }+1 \right) Y+2q=0\) are perpendicular to a common line for
- (a)
exactly one value of p
- (b)
exactly two values of p
- (c)
more than two values of p
- (d)
no value of p
The line parallel to the X-axis and passing through the intersection of the lines aX + 2bY - 3a = 0 and bX - 2aY - 3a = 0, where \(\left( a,b \right) \neq \left( 0,0 \right) \) is
- (a)
above the X-axis at a distance of (2/3) from it.
- (b)
above the X-axis at a distance of (3/2) from it.
- (c)
below the X-axis at a distance of (2/3) from it.
- (d)
below the X-axis at a distance of (3/2) from it.
If non-zero numbers a,b and c are in HP, then the straight line \(\frac { X }{ a } +\frac { Y }{ b } +\frac { 1 }{ c } =0\) always passes through a fixed point. That point is
- (a)
\(\left( 1,-\frac { 1 }{ 2 } \right) \)
- (b)
(1, - 2)
- (c)
(-1, -2)
- (d)
(-1, 2)
Let A (2, - 3) and B (- 2, 1) be the vertices of a \(\triangle ABC\). If the centroid of this triangle moves on the line 2X + 3Y = 1, then the locus of the vertex C is the line
- (a)
2x + 3y = 9
- (b)
2x - 3y = 7
- (c)
3x + 2y = 5
- (d)
3x - 2y = 3