JEE Main Mathematics - Cartesian Coordinate System
Exam Duration: 60 Mins Total Questions : 30
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 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)
\(p\)
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
The equation of the line passing through (- 3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6) is
- (a)
5x - y + 20 = 0
- (b)
x - 5y + 20 = 0
- (c)
x - y + 20 = 0
- (d)
5x + 5y + 20 = 0
The angle between the lines \(\sqrt{3}\)X + Y = 1 and X + \(\sqrt{3}\)Y = 1 is equal to
- (a)
30°
- (b)
60°
- (c)
90°
- (d)
45°
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
A line which makes an acute angle \(\theta \) with the positive direction of X-axis is drawn through the point P(3, 4) to meet the line X = 6 at R and Y = 8 at S, then
- (a)
\(PS=4\sin { \theta } \)
- (b)
PR = 3 cosec \(\theta \)
- (c)
\(PR+PS=\frac { 2\left( 3\sin { \theta -4\cos { \theta } } \right) }{ \sin { 2\theta } } \)
- (d)
\(\frac { 9 }{ { \left( PR \right) }^{ 2 } } +\frac { 16 }{ { \left( PS \right) }^{ 2 } } =1\)
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)
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