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
The base of a triangle lies along the line X = a and is of length a. The area of the triangle is a2 . If the vertex lies on the line parallel to the base of triangle, then that equation of line is
- (a)
x = 0
- (b)
x = a
- (c)
x = 3a
- (d)
x = - 3a
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 foot of the perpendicular from (2, 3) upon the line 4X - 5Y + 8 = 0 is
- (a)
(0, 0)
- (b)
(1, 1)
- (c)
\(\left( \frac { 41 }{ 78 } ,\frac { 128 }{ 75 } \right) \)
- (d)
\(\left( \frac { 78 }{ 41 } ,\frac { 128 }{ 41 } \right) \)
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)
\(p\)
A straight line L with negative slope passes through the point (8, 2) and cuts the positive coordinates axes at points P and Q. As L varies, the absolute minimum value of OP + OQ is (O is origin)
- (a)
10
- (b)
18
- (c)
16
- (d)
112
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 a family of lines a(2X + Y + 4) + b(X - 2Y - 3) = 0. Then, the number of lines belonging to the family at a distance of \(\sqrt{10}\) from P(2, - 3) is
- (a)
0
- (b)
1
- (c)
2
- (d)
4
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)
\(a+b+c\neq 0\)
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 line L given by \(\frac { X }{ 5 } +\frac { Y }{ b } =1\) passes through the point (13, 32). The line K is parallel to L and has the equation \(\frac { X }{ c } +\frac { Y }{ 3 } =1\) Then, the distance between L and K is
- (a)
\(\frac { 23 }{ \sqrt { 15 } } \)
- (b)
\(\sqrt{17}\)
- (c)
\(\frac { 17 }{ \sqrt { 15 } } \)
- (d)
\(\frac { 23 }{ \sqrt { 17 } } \)
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.
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