IISER Mathematics - Coordinate Geometry - I
Exam Duration: 45 Mins Total Questions : 30
The points \((0, \frac 8{3})\) (1, 3) and (82, 30) are the vertices of
- (a)
obtuse angled triangle
- (b)
acute angled triangle
- (c)
right angled triangle
- (d)
None of these
Let A\((0.\frac{8}{3})\), B(1,3), C(82,30)
Slope of AB=\(\frac{3-8/3}{1-0}=\frac{1}{3}\)
Slope of BC=\(\frac{30-3}{82-1}=\frac{27}{81}=\frac{1}{3}\)
Slope of AB = Slope of BC
Thus the given points are collinear
Hence, the correct alternative is (d)
Let 0<\(\alpha\)<\(\frac { \pi }{ 2 } \) be a fixed angle. If P(cos\(\theta \), sin\(\theta \)) and Q (cos(\(\alpha - \theta\)), sin(\(\alpha - \theta\))), then Q is obtained from P by
- (a)
clock wise raotaion around the origin through an angle \(\alpha\)
- (b)
anticlockwise rotation around the origin
- (c)
reflection in the line through the origin with sope tan \(\alpha\)
- (d)
reflection in the line through the origin with slope tan \((\frac {\alpha}{2})\)
If (x, y) are coordinates of the point (x, y) when reflected through the line y = x tan \(\alpha\) , then
\(\left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} cos2\theta & sin2\theta \\ sin2\theta & -cos2\theta \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) \)
Taking \(\theta =\frac { \alpha }{ 2 } or\quad 2\theta =\alpha \quad and\quad x=cos\theta ,\quad y=sin\theta ,\quad we\quad get\)
\(\left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} cos\alpha & sin\alpha \\ sin\alpha & -cos\alpha \end{matrix} \right) \left( \begin{matrix} cos\theta \\ sin\theta \end{matrix} \right) \)
\(\left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} cos\alpha cos\theta +sin\alpha sin\theta \\ sin\alpha cos\theta -cos\alpha sin\theta \end{matrix} \right) \)
\(=\left( \begin{matrix} cos(\alpha -\theta ) \\ sin(\alpha -\theta ) \end{matrix} \right) \)
Thus P(cos \(\theta \), sin \(\theta \)) is transformed to Q(cos (\(\alpha \) - \(\theta \)), sin(\(\alpha \) - \(\theta \)))
Hence. the correct alternative is (d).
Area of the triangle formed by the llines y=m1x+c1, y=m2x+c2, x=0 is
- (a)
\(\begin{vmatrix} \frac { { c }_{ 2 }-{ c }_{ 1 } }{ ({ m }_{ 2 }-{ m }_{ 2 })^2 } \end{vmatrix}\)
- (b)
\(\frac { 1 }{ 2 } \begin{vmatrix} \frac { ({ c }_{ 2 }-{ c }_{ 1 })^{ 2 } }{ ({ m }_{ 2 }-{ m }_{ 2 }) } \end{vmatrix}\)
- (c)
\(\frac { 1 }{ |m_2-m_1|^2 } \)
- (d)
None of these
Area of the triangle formed by the line
y=m1x+c1,y=m2x+c2,y=m3x+c3, is given by
=\(\frac { 1 }{ 2 } \left| \frac { { (c }_{ 2 }-{ c }_{ 1 }{ ) }^{ 2 } }{ ({ m }_{ 2 }-{ m }_{ 1 }) } +\frac { ({ c }_{ 3 }-{ c }_{ 2 }) }{ { (m }_{ 3 }-{ m }_{ 2 }) } +\frac { { (c }_{ 1 }-{ c }_{ 3 }{ ) }^{ 2 } }{ ({ m }_{ 1 }-{ m }_{ 3 }) } \right| \)
Here m3=∞
\(\therefore\) Area of the triangle=\(\frac { 1 }{ 2 } \left| \frac { { (c }_{ 2 }-{ c }_{ 1 }{ ) }^{ 2 } }{ ({ m }_{ 2 }-{ m }_{ 1 }) } \right| \)
Hence the correct alternative is (a).
The area of triangle is 5. Two of its vertices are (2, 1) and (-3, -2). The third vertex lies on the line y-x=3. Then the thrid vertex is at
- (a)
\((\frac 7{2},\frac {13}{2} )\)
- (b)
\((-\frac 3{2},\frac {3}{2} )\)
- (c)
\((-\frac 7{2},\frac {13}{2} )\)
- (d)
\((\frac 3{2},\frac {3}{2} )\)
Let the third vertex be(x, y)
\(\therefore \frac { 1 }{ 2 } \left| \begin{matrix} 2 & 1 & 1 \\ 3 & -2 & 1 \\ x & y & 1 \end{matrix} \right| \)
=\(\frac{1}{2}\)[2(-2-y)-1(3-x)+1(3y+2x)]=5
or 3x+y=17 ......(i)
Also -x+y=3 ....(ii)
Solving (i)and (ii) we getx=\(\frac{7}{2}\),y=\(\frac{13}{2}\)
The coordinates of the image point of (1, 1) in the line 2x+3y+5=0, are
- (a)
\((\frac {27}{13}, \frac {47}{13})\)
- (b)
\((\frac {-47}{13}, \frac {27}{13})\)
- (c)
\((\frac {-27}{13}, \frac {-47}{13})\)
- (d)
\((\frac {27}{13}, \frac {47}{13})\)
\(\frac { h-{ x }_{ 1 } }{ a } =\frac { k-{ y }_{ 1 } }{ b } =\frac { -2(a{ x }_{ 1 }+b{ y }_{ 1 }+c) }{ { a }^{ 2 }+{ b }^{ 2 } } \)
Here, x1=1,y1=1,a=2,b=3,c=5
\(\frac{h-2}{2}=\frac{k-3}{3}=-\frac{-2(2\times1+3\times1+5)}{2^2+3^2}=\frac{-20}{13}\)
h=\(-\frac{27}{13}\),k=\(-\frac{47}{13}\)
Hence the correct alternative is (c).
The algebraic sum of the perpendicular distances of the points (1,0), (0,1), (2,2) from a variable line is zero; then the line passes through a fixed point
- (a)
(1, 1)
- (b)
(2, 3)
- (c)
(2, 4)
- (d)
(4, 3)
Required point is
\((\frac{1+0+2}{3},\frac{0+1+2}{3})\)
i.e. (1,1)
Hence the correct alternative is (a).
The equation 12x2+7xy-py2-18x+qy+6=0, represents a pair of perpendicular straight lines, if
- (a)
p = 12, q = 1
- (b)
p = 1, q = 12
- (c)
p = -1, q = 12
- (d)
p = 1, q = -12
The equation of the pair of straight lines joining the points of intersection of the line y = 3x+2 and the curve x2+2xy-3y2+4x+8y-11=0 with origin, is
- (a)
7x2+2xy-y2=0
- (b)
7x2-2xy+y2=0
- (c)
7x2-2xy-y2=0
- (d)
NONE OF THESE
Re-writing equation of the line as
\(\frac{y-3x}{2}=1\) ...(i)
Re-writing the equation of the curve as
x2+2xy-3y2+4x.1+8y.1-11.12=0 ...(ii)
Putting value of 1 from (i) in (ii) to make (ii) equation homogeneous, we get
x2+2xy-3y2+4x.\((\frac{y-3x}{2})\)+8y\((\frac{y-3x}{2})\)-11.\((\frac{y-3x}{2})^2\)=0
which on simplification gives
7x2 - 2xy - y2=0
the required equation of the pair of lines.
Hence, the correct alternative (c).
The equation of the circle which cuts an intercept of 8 units on the x-axis and touches the y-axis at(0, 3)
- (a)
\({ x }^{ 2 }+{ y }^{ 2 }-10x\pm 6y+9=0\)
- (b)
\({ x }^{ 2 }+{ y }^{ 2 }+10x\pm 6y+9=0\)
- (c)
\({ x }^{ 2 }+{ y }^{ 2 }\pm 10x-6y+9=0\\ \)
- (d)
NONE OF THESE
Let the circle be
x2+y2+2gx+2fy+c=0 ....(i)
Intercept made by (i) on the x-axis
2\(\sqrt{g^2-c}=8\)
g2-c=16 ....(ii)
Circle (i) touches y-axis at (0, 3).
ஃy2 + 2fy+c=0,
which must have each root equal to 3.
Sum of roots = -2 f = 3+3
⇒f=-3
Product of roots=c=3x3
⇒c=9
from(ii)
x2+y2土10x-6y+9=0
Hence the correct alternative is (c).
The maximum number of points with rational coordiantes on a circle whose centre is \((\sqrt { 3 } ,0)\) is
- (a)
one
- (b)
two
- (c)
four
- (d)
infintely many
Let r be the radius of the circle, then its equation is
(x-\(\sqrt3\))2+y2=r2
x2+y2-2\(\sqrt3\)x=r2-3
x2+y2-r2-3=2\(\sqrt3\)x
If x,y rational, then
2\(\sqrt3\)x=0⇒x=0
ஃy2-r2+3=0
which is quadratic in y giving two rational values of y
Thus, there can be at the most two points, with rational.
Correct alternative is (b).
The distance of the point (1,2) from the radical axes of the circles x2+y2+6x-16=0 and x2+y2-2x-6y-6=0 is
- (a)
1
- (b)
2
- (c)
3
- (d)
4
Equation of the radical axis is
(x2 + y2 + 6x -16) - (x2 + y2 - 2x -6y - 6) = 0
or 8x + 6y -10=0
or 4x+3y -5=0
Distance of the point (l, 2) from (i)
\(=\left| \frac { 4.1+3.2-5 }{ \sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 } } } \right| =\frac { 5 }{ 5 } =1\)
If the two circles x2+y2=r2 and (x-5)2+y2=9 intersect in two distinct points,then
- (a)
2
- (b)
r<2
- (c)
r=2
- (d)
r>2
If C1, C2 and r1 are the centres and radii of the two circles respectively, then these circles intersect, if C1C2<r1+r2 and C1C2<r1-r2
Now C1(0, 0), C2(5, 0) and r1= r, r2= 3
ஃ 5<r+3 or 2<r
Also 5 > r -3 or 8 > r
ஃ 2<r<8
The angle between the tangents from the point (4,3) to the circle x2+y2-2x-4y=0 is
- (a)
300
- (b)
450
- (c)
600
- (d)
900
Combined equation of the two tangents is given by
SS1=T2
i.e (x2 + y2 -2x -4y)(42+32-2x4-4x3)
=[4x+3y-(x+4)-2(y+3)]2
i.e. 5(x2+y2-2x-4y)=(3x+y-10)2
or 4x2+6xy-4y2-50x+100=0 ....(i)
Since, Coefficient of x2 + coefficient of y2 = 0
Thus, the lines represented by (i) are perpendicular.
The radical center of the circles x2+y2+4x+7=0, 2x2+2y2+3x+5y+9=0, and x2+y2+y=0, is
- (a)
(-2,1)
- (b)
(-2,-1)
- (c)
(2,-1)
- (d)
(2,1)
The radical axis of the first two circles is
(x2+y2+4x+7)-(x2+y2+ \(\frac{3}{2}\)x+\(\frac{5}{2}\)y+\(\frac{9}{2}\))=0
or x -y +1=0 ....(i)
The radical axis of the first and the third circles is
(x2+y2+4x+7)-(x2+y2+y)=0
or 4x-y+7=0 .....(ii)
Solving (i) and (ii), we get the coordinate axes and centre of circle is (-2, -1).
The length of the tangent from the point (5, 1) to the circle x2+y2+6x-4y-3=0 is
- (a)
81
- (b)
29
- (c)
7
- (d)
NONE OF THESE
Length of tangent
=\(\sqrt{5^2+1^2+6\times5-4\times1-3}\)
[∵PT=\(\sqrt { { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+2gx_{ 1 }+2gy_{ 1 }+c } \)]
=\(\sqrt{56-7}=7\)
The equation of the tangents drawn from the origin to the circle
x2+y2-2rx-2hy+h2=0 are
- (a)
x=0
- (b)
y=0
- (c)
(h2-r2)x-2rhy=0, x=0
- (d)
(h2-r2)x+2rhy=0, y=0
Equations of tangents are given by
SS1=T2
or (x2+y2-2rx-2hy+h2)
=(x.0+y.0-r(x+0)-h(y+0)+h2)2
or (x2+y2-2rx-2hy+h2)(+h2)
or (x2+y2-2rn-2hy+h2)(+h2)
=(rx+hy-h2)2
or h2(x2+y2-2rx-2hy+h2)
=r2x2+h2y2+h4+2rhxy-2h3y-2rh2x
or (h2-r2)x2-2rhxy=0 (on Simplification)
or [(h2-r2)x-2rhy]x=0
Therefore equations of the tangents are
(h2-r2)x-2rhy=0 and x = 0
Hence, the correct alternative is (c).
The image of P (a, b) on y = -' X is Q and the image of Q on the line y = X is R. Then the mid point of R is
- (a)
(a + b, b + a)
- (b)
\(\left( \frac { a+b }{ 2 } ,\frac { b+a }{ 2 } \right) \)
- (c)
(a-b,b+a)
- (d)
(0,0)
The image of P (a, b) on y = - X is Q (-b, - a) (interchan~e and change signs) and the image of Q (-b, -a) on y = X IS R (-a, - b)(merely interchange). :. The mid point of PR is (0, 0).
The line 3x - 4y + 7 = 0 is rotated through an angle \(\frac { \pi }{ 4 } \) in the clockwise direction about the point (-1,1). The equation of the line in its new position is
- (a)
7y + x - 6 = 0
- (b)
7y - x - 6 = 0
- (c)
7y + x + 6 = 0
- (d)
7y - x + 6 = 0
As(-1,1) is a point on 3x-4y+7=0 the rotation is possible
Slope of the line = \(\frac { 3 }{ 4 } \)
Slope of the line in its new position = \(\frac { \frac { 3 }{ 4 } -1 }{ 1+\frac { 3 }{ 4 } } =-\frac { 1 }{ 7 } \)
The required equation is y-1=\(-\frac { 1 }{ 7 } (x+1)\)
7y+x-6=0
Given a family of lines a (2x + y + 4) + b (x - 2y - 3) = 0, the number of lines belonging to the family at a distance -10 from P (2, - 3) is
- (a)
0
- (b)
1
- (c)
2
- (d)
4
P= \(\left| \frac { a(4-3+4)+b(2+6-3) }{ (2a+b)^{ 2 }+(a-2b)^{ 2 } } \right| =\sqrt { 10 } \)
= \(25(a+b)^{ 2 }=10(5a^{ 2 }+5b^{ 2 })\)
\(\Rightarrow 25(a+b)^{ 2 }=0\Rightarrow a=b\)
only line is 3x-y+1=0
If Sa + 4b + 20c = t, then the value of t for which the line ax + by + c- 1 = 0 always passes through a fixed point is
- (a)
0
- (b)
20
- (c)
30
- (d)
none of these
Equation of line \(\frac { ax }{ c-1 } +\frac { by }{ c-1 } +1=0\) has two independent parameters. It can pass through a fixed point if it contains only one independent parameter. Now there must be one relation \(\frac { a }{ c-1 } ,\qquad \frac { b }{ c-1 } \) independent of a, band c so that \(\frac { a }{ c-1 } \) can be expressed in terms of \(\frac { b }{ c-1 } \) and straight line contains only one independent parameter. Now that given relation can be expressed as \(\frac { 5a }{ c-1 } +\frac { 4b }{ c-1 } =\frac { t-20c }{ c-1 } \) RHS is independent of c if t = 20.
The straight line y = x - 2 rotates about a point where it cuts x-axis and becomes perpendicular on the straight line ax + by + c = 0, then its equation is
- (a)
ax + by + 2a=0
- (b)
ay - bx + 2b=0
- (c)
ax + by + 2b=0
- (d)
none of these
Equation of line passing through (2,0) and perpendicular to ax+by+c=0
Then required equation is y-0 = \(\frac { b }{ a } (x-2)\)
ay=bx-2b
ay-bx+2b=0
The diagonals of the parallelogram whose sides are Ix + my + n = 0, Ix + my + n' = 0, mx + ly + n = 0, mx + ly + n' = °include an angle
- (a)
\(\pi /3\)
- (b)
\(\pi /2\)
- (c)
\(tan^{ -1 }\left( \frac { l^{ 2 }-m^{ 2 } }{ l^{ 2 }+m^{ 2 } } \right) \)
- (d)
\(tan^{ -1 }\left( \frac { 2lm }{ l^{ 2 }+m^{ 2 } } \right) \)
Since Distance between parallel lines are same Parallelogram is a rhombus
In rhombus diagonals are perpendicular to each other. Hence angle between diagonals = \(\pi /2\)
The position of a moving point in the xy-plane at time t is given by \(\left( ucos\alpha \quad t,\quad u\quad sin\quad \alpha t-\frac { 1 }{ 2 } gt^{ 2 } \right) \)where u,a, g are constants. The locus of the moving point is
- (a)
a circle
- (b)
a parabola
- (c)
an ellipse
- (d)
none of these
Let x=u cos \(\alpha .t\)
and y= u sin \(\alpha .t\) - \(\frac { 1 }{ 2 } gt^{ 2 }\)
Substituting the value of t from Eq (i) in (ii) then
\(y=x\quad tan\alpha -\frac { 1 }{ 2 } \frac { gx^{ 2 } }{ u^{ 2 }cos^{ 2 }x } \)
It is a parabola
If (- 6, - 4), (3, 5), (-2, 1) are the vertices of a parallelogram, then remaining vertex cannot be
- (a)
(0,-1)
- (b)
(-1,0)
- (c)
(-11, -8)
- (d)
(7, 10)
If remaining vertex is (a,b) then
\(\frac { \alpha +3 }{ 2 } =-4\frac { \beta +5 }{ 2 } =-3/2\)
\(\alpha =-11,\beta =-8\\ \frac { \alpha -6 }{ 2 } =\frac { 3-2 }{ 2 } ,\frac { \beta -4 }{ 2 } =\frac { 5+1 }{ 2 } \)
\(\alpha =7,\beta =10\\ \frac { \alpha -2 }{ 2 } =\frac { 3-6 }{ 2 } ,\frac { \beta +1 }{ 2 } =\frac { 5-4 }{ 2 } \)
\(\alpha =-1,\beta =0\)
Possibilities of remaining vertex are
(-11,-8)or (7,10) or (-1,0)
If 3a + 2b + 6c = 0, the family of straight lines ax + by + c = 0 passes through a fixed point whose coordinates are given by
- (a)
(1/2, 1/3)
- (b)
(2,3)
- (c)
(3, 2)
- (d)
(1/3, 1/2)
3a+2b+6c=0
c=\(\frac { (3a+3b) }{ 6 } \)
ax+by+c=0
\(\Rightarrow ax+by-\frac { (3a+2b) }{ 6 } =0\)
\(\Rightarrow 3a(2x-1)+2b(3y-1)=0\)
\(\Rightarrow (2x-1)+\frac { 2b }{ 3a } (3y-1)=0\)
\(p+\lambda Q=0\therefore p=0,Q=0\)
Then 2x-1=0,3y-1=0;x=1/2,y=-1/3
Hence fixed point is \(\left( \frac { 1 }{ 2 } ,\frac { 1 }{ 3 } \right) \)
If P (1, 0), Q (-1, 0) and R (2, 0) are three given points, then the locus of points satisfying the relation \((SQ)^{ 2 }+(SR)^{ 2 }=2(SP)^{ 2 }\)
- (a)
a straight line parallel to x-axis
- (b)
circle through origin
- (c)
circle with centre at the origin
- (d)
a straight line parallel to y-axis
Let S=(x,y)
given \((SQ)^{ 2 }+(SR)^{ 2 }=2(SP)^{ 2 }\)
\(\Rightarrow (x+1)^{ 2 }+y^{ 2 }+(x-2)^{ 2 }+y^{ 2 }-2(x-1)^{ 2 }+y^{ 2 })\)
\(\Rightarrow 2x^{ 2 }+2y^{ 2 }-2x+5-2(x^{ 2 }+y^{ 2 }-2x+)\)
\(\Rightarrow 2x+3=0\Rightarrow x=-\frac { 3 }{ 2 } \)
The coordinates of the point P on the line 2x + 3y + 1 = 0, such that IPA - PB I is maximum, where A is (2, 0) and B is (0,2) is
- (a)
(5, - 3)
- (b)
(7, - 5)
- (c)
(9, - 7)
- (d)
(11, - 9)
|PA-PB|<|AB|
M<aximum value of |PA-PB| is |AB| which is possible only when P,A,B are collinear
if P(x,y) then equation AB is
\(\frac { x }{ 2 } +\frac { y }{ 2 } =1\Rightarrow x+y=2\)
Now solving Eq (1)
and 2x+3y+1=0
There we get
x=7,y=-5
P=(7,-5)
Consider the straight lines x + 2y + 4 = 0 and 4x + 2y - 1 = O.The line 6x + 6y + 7 = 0 is
- (a)
bisector of the angle including origin
- (b)
bisector of acute angle
- (c)
bisector of obtuse angle
- (d)
none of the above
x+2y+4=0 and 4x+2y-1=0
x+2y+4=0 and -4x-2y+4+1=0
Here (1)(-4)+(2)(-2)=-8<0
Bisector of the angle including the acure angle bisector and origin is
\(\frac { x+2y+4 }{ \sqrt { 5 } } =\frac { (-4x-2y+1) }{ 2\sqrt { 5 } } \)
6x+6y+7=0
Consider the equation y - Yl = m (x - x.). If m and Xl are fixed and different lines are drawn for different values of y1, then
- (a)
the lines will pass through a fixed point
- (b)
there will be a set of parallel lines
- (c)
all the lines intersect the line x = x1
- (d)
all the lines will be parallel to the line y = X1
\((y-y_{ 1 })-m(x-x_{ 1 })0\)
is family of lines
\(y-y_{ 1 }=0,x-x_{ 1 }=0\)
Then, \(y-y_{ 1 }\quad \& \quad x-x_{ 1 }\)
The family of lines \((2cos\theta +3sin\theta )x+(3cos\theta -5sin\theta )y\)for different values of a and b is concurrent at \((x_{ 2 },y_{ 2 })\) then
- (a)
\(x_{ 2 }+y_{ 2 }=x_{ 1 }\)
- (b)
\(x_{ 2 }+y_{ 2 }=y_{ 1 }\)
- (c)
\(6x_{ 2 }+y_{ 2 }=2x_{ 1 }+y_{ 1 }\)
- (d)
none of these
Given family of lines are
\(cos\theta (2x+3y-5)+sin\theta (3x-5y+2)=0\)
\(a(x+y-1)+b(2x-3y+1)=0\)
2x+3y-5=0
3x-5y+2=0
x+y-1=0
2x-3y+1=0
\(\Rightarrow x_{ 1 }=1,y_{ 1 }=1,x_{ 2 }=\frac { 2 }{ 5 } ,y_{ 2 }=\frac { 3 }{ 5 } \)