Eamcet Mathematics - Cartesian Coordinate System Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 20
The two lines aX + bY = c and \({ a }^{ \prime }X+{ b }^{ \prime }Y={ c }^{ \prime }\) are perpendicular, if
- (a)
\(a{ a }^{ \prime }+{ bb }^{ \prime }=0\)
- (b)
\(a{ b }^{ \prime }={ ba }^{ \prime }\)
- (c)
\(ab+{ a }^{ \prime }{ b }^{ \prime }=0\)
- (d)
\(a{ b }^{ \prime }+{ ba }^{ \prime }=0\)
Slope of the line aX + bY = c is \(\frac { -a }{ b } \) and the slope of the line \({ a }^{ \prime }X+{ b }^{ \prime }Y={ C }^{ \prime }\) is \(\frac { -{ a }^{ \prime } }{ { b }^{ \prime } } \)
The lines are perpendicular, if \({ m }_{ 1 }.{ m }_{ 2 }=-1\)
\(\therefore \left( \frac { -a }{ b } \right) \left( \frac { -{ a }^{ \prime } }{ { b }^{ \prime } } \right) =-1\)
\(\Rightarrow a{ a }^{ \prime }+b{ b }^{ \prime }=0\)
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
Let the slope of the required line be m. Again, this line is perpendicular to the line X + Y + 7 = 0. Therefore, we have m = - (- 1) = 1
Hence, the required equation of the line passing through (1, 2) is
\(Y-2=X-1\Rightarrow Y-X-1=0\)
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 family of lines parallel to 3X + 4Y + 5 = 0 is 3X + 4y + k = 0 because after comparing the given equation, we will replace 5 from k.
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
Let h be the height of the triangle. Since, the area of the triangle is a2
\(\therefore \frac { 1 }{ 2 } \times a\times h={ a }^{ 2 }\Rightarrow h=2a\)
Since, the base lies along the line X = a, the vertex lies on the line parallel to the base of triangle at a distance 2a from it. So, the required lines are
\(X=a\pm 2a\Rightarrow X=-a,X=3a\)
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)
\(\left( \frac { 78 }{ 41 } ,\frac { 128 }{ 41 } \right) \)
If foot of perpendicular is (h,k), then
\(\frac { h-2 }{ 4 } =\frac { k-3 }{ -5 } =\frac { -\left( 8-15+8) \right) }{ 16+25 } =\frac { -1 }{ 41 } \)
\(\Rightarrow h=2-\frac { 4 }{ 41 } =\frac { 78 }{ 41 } \)
and \(k=3+\frac { 5 }{ 41 } =\frac { 128 }{ 41 } \)
So, the required point is \(\left( \frac { 78 }{ 41 } ,\frac { 128 }{ 41 } \right) \)
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
The equation of the line L bey y - 2 = m(x-8), m<0
Coordinates of P and Q are \(\left( 8-\frac { 2 }{ m } ,0 \right) \) respectively.
so, OP + OQ = 8 - \(2\over m\) + 2 - 8m = 10 + \(2\over(-m)\) + 8(-m)
≥10 + 2 \(\ge 10+2\sqrt { \frac { 2 }{ (-m) } \times 8(-m) } \ge 18\)
[since, AM \(\ge\) GM for two positive numbers]
so, absolute minimum value Of OP + OQ is 18.
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
\(P=\left| \frac { a\left( 4-3+4 \right) +b\left( 2+6-3 \right) }{ \sqrt { { \left( 2a+b \right) }^{ 2 }+{ \left( a-2b \right) }^{ 2 } } } \right| =\sqrt { 10 } \)
\(\Rightarrow \quad 25{ \left( a+b \right) }^{ 2 }=10\left( { 5a }^{ 2 }+{ 5b }^{ 2 } \right) \)
\(\Rightarrow \quad { \left( a-b \right) }^{ 2 }=0\Rightarrow a=b\)
So, only line is 3X - Y + 1 = 0.
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
Let the point p and Q be (2, 5) and (-3, 6)
Let slope of AB be m.
Since, AB\(\bot \)PQ
ஃ Slope of AB x Slope of PQ = -1
⇒ m x \({Y_2-y_1\over x_2-x_1}=-1\Rightarrow m\times{6-5\over-3-2}=-1\)
⇒ m x \(1\over-5\) = -1 ⇒ m = 5
So, equation of line AB
⇒ y - 5 = 5(x+3)
⇒ y - 5 = 5x + 15
ஃ 5x - y + 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°
Given equation of lines are \(\sqrt3 x+y=1\) and x + \(\sqrt{3}\)y = 1
i.e., y = -\(\sqrt{3}\)x + 1
and y = \(-1\over\sqrt3\) x + \(1\over\sqrt3\)
On comparing with y = m1x + c1 and y = m2x + c2, we get
m1 = -\(\sqrt{3}\) and m2 = -\(1\over\sqrt3\)
ஃ tan\(\theta\) = \(\left| \frac { -\sqrt { 3 } -\left( -\frac { 1 }{ \sqrt { 3 } } \right) }{ 1+(-\sqrt { 3 } )-\left( -\frac { 1 }{ \sqrt { 3 } } \right) } \right| \left[ \because \quad tan\theta =\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \right] \)
\(\quad tan\theta =\left| \frac { -\sqrt { 3 } +\frac { 1 }{ \sqrt { 3 } } }{ 1+1 } \right| =\left| \frac { \frac { -3+1 }{ \sqrt { 3 } } }{ 2 } \right| \)
⇒ tan\(\theta\) = \(1\over\sqrt3\) ⇒ tan\(\theta\) - tan 30° ⇒ \(\theta\) = 30°
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
Given equation is \({x\over a}+{y\over b}=1\)
At origin, perpendicular distance from line (i) is
p = \(\left| \frac { 0+0-1 }{ \sqrt { \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } } } \right| \Rightarrow p=\frac { 1 }{ \sqrt { \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } } } \)
\(\Rightarrow \frac { 1 }{ { p }^{ 2 } } =\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } \)
Since, a2, p2 and b2 are in AP
ஃ 2p2 = a2 + b2
Now, \(\Rightarrow \quad \frac { 1 }{ { p }^{ 2 } } =\frac { { a }^{ 2 }+{ b }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } \Rightarrow { p }^{ 2 }=\frac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }{ +b }^{ 2 } } \)
⇒ \(\frac { 1 }{ { p }^{ 2 } } =\frac { { a }^{ 2 }+{ b }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } \Rightarrow { p }^{ 2 }=\frac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } } \)
⇒ (a2 + b2) = 2a2b2
⇒ a4 + b4 = 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
The point (3, 0) does not lie on the diagonal X = 2Y.
Let the equation of a side through the vertex (3, 0) be Y = mX + c.Since, the angle between a side and a diagonal of a square is \(\\ \frac { \pi }{ 4 } \) .
We have, \(\tan { \frac { \pi }{ 4 } } =\frac { m-\frac { 1 }{ 2 } }{ 1+m\left( \frac { 1 }{ 2 } \right) } =\frac { 2m-1 }{ 2+m } \quad \left[ \because Y=\frac { 1 }{ 2 } X \right] \)
\(\Rightarrow \quad \quad m=3\)
Thus, the equation of a side through (3, 0) is Y = 3 (X - 3).
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)
\(-2x+y\ge 0\)
Clearly, points (1, 3), (5, 0) and (- 1, 2) satisfy the inequalities
\(3x+2y\ge 0\) and \(2x-3y-12\le 0\).
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)
\(a+b+c\neq 0\)
The equation represent same line, if
\(\frac { { b }^{ 3 }-{ c }^{ 3 } }{ b-c } =\frac { { c }^{ 3 }-{ a }^{ 3 } }{ c-a } =\frac { { a }^{ 3 }-{ b }^{ 3 } }{ a-b } =k\)
Clearly, we get \({ b }^{ 3 }-{ c }^{ 3 }=k\left( b-c \right) \)
\({ c }^{ 3 }-{ a }^{ 3 }=k\left( c-a \right) \)
\({ a }^{ 3 }-{ b }^{ 3 }=k\left( a-b \right) \)
\(\Rightarrow \quad b-c=0\quad or\quad { b }^{ 2 }+{ c }^{ 2 }+bc=k\quad \quad \quad ...(i)\)
\(\Rightarrow \quad c-a=0\quad or\quad { c }^{ 2 }+{ a }^{ 2 }+ca=k\quad \quad \quad ...(ii)\)
\(\Rightarrow \quad a-b=0\quad or\quad { a }^{ 2 }+b^{ 2 }+ab=k\quad \quad \quad ...(iii)\)
On solving Eqs. (i), (ii) and (iii), we get a = b = c
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)
\(\frac { 2 }{ 1|9| } +\frac { 2 }{ 3|7| } +\frac { 1 }{ 5|5| } =\frac { 2^{ n } }{ n! } \)
\(\Rightarrow \frac { 1 }{ 10! } \left[ \frac { 2\times 10! }{ 1|9| } +\frac { 2\times 10! }{ 3!7! } +\frac { 10! }{ 5!5! } \right] =\frac { 2^{ m } }{ n! } \)
\(\Rightarrow \frac { 1 }{ 10! } \left\{ 2\times 10C_{ 1 }+2\times 10C_{ 3 }+10C_{ 5 } \right\} =\frac { 2^{ m } }{ n! } \)
\(\Rightarrow \frac { 1 }{ 10! } \left\{ 2\times 10C_{ 1 }+2\times 10C_{ 3 }+10C_{ 5 }+10C_{ 7 }+10C_{ 9 } \right\} =\frac { 2^{ m } }{ n! } \)
\(\left[ \because ^{ n }C_{ t }=^{ n }C_{ n }..t \right] \)
\(\Rightarrow \frac { 1 }{ 10 } \left( 2 \right) ^{ 10^{ -1 } }=\frac { 2^{ m } }{ n! } \Rightarrow \) m=9 and n=10
Also, x -y + =0 and x + y + 3 = .0 are perpendicular to each other. So, orthocentre is the point of intersection which is -(-2, 1).
\(\therefore \) -2=2m-2n\(\Rightarrow \) -1=m-n
Hence point is (2m,-2n,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
Given, lines are 4aX + 2aY + c = 0 ....(i)
and 5bX + 2bY + d = 0 ....(ii)
Let the point of intersection be (h, -h).
On substituting it in Eqs. (i) and (ii), we get
4ah - 2ah + c = 0 and 5bh - 2bh + d = 0
On eliminating h in the above two equations, we get
\(\frac { -c }{ 2a } =\frac { -d }{ 3b } \)
Hence, 3bc - 2ad = 0
The lines \(X+Y=\left| a \right| \) and \(aX-Y=1\) intersect each other in the first quadrant. Then, the set of all possible values of a in the interval.
- (a)
(- 1, 1]
- (b)
\(\left( 0,\infty \right) \)
- (c)
\([1,\infty )\)
- (d)
\(\left( -1,\infty \right) \)
As x+y=|a| and ax-y =1 intersect in I quadrant
Thus, x and y-intercept are positive.
\(\therefore x=\frac { 1+|a| }{ 1+a } \ge \) and y=\(\therefore x=\frac { a|a|-1 }{ a+1 } \ge 0\)
\(\Rightarrow 1+a\ge 0\) and \(y=\frac { a|a|-1 }{ a+1 } \ge 0\)
\(\Rightarrow a\ge -1\) and a|a|-1 \(\ge \) 0
If -1<a<0 and a2 \(\ge 1\) ...(i)
\(\Rightarrow \) -a2 \(\ge \) 1 [not possible]
if \(a\ge 0\) and \(a^{ 2 }\ge 1\)
\(\Rightarrow a\ge 1\) (ii)
\(\therefore a\ge 1\) or \(a\in \left[ 1,\infty \right] \)
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
Lines perpendicular to same line are parallel to each other.
\(\therefore \quad \quad \quad -p\left( { p }^{ 2 }+1 \right) ={ p }^{ 2 }+1\\ \Rightarrow \quad \quad \quad p=-1\)
Hence, there is exactly one 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.
Equation of a line passing through the intersection of lines
ax + 2by + 3b = 0 and bx - 2ay - 3a= 0 is
(ax + 2by + 3b) + λ(bx - 2ay -3a)=0 .....(i)
Now, this line is parallel to X-axis, so coefficient of x should be zero.
i.e., a+λb=0
⇒ λ=-\(\frac{a}{b}\)
On putting this value in Eq.(i) we get
b (ax + 2by + 3b) - a (bx - 2ay - 3a)=0
⇒ 2b2y+3b2+2a2y+3a 2=0
⇒ 2(b2+a2)+3(b2+a2 )=0
⇒ y=\(-\frac{3}{2}\)
The negative sign shows that the line is below X-axis at a distance \(\frac{3}{2}\) from it.
Alternate Method
Equations oi given lines are
ax+2by+3b=0 ....(i)
and bx-2ay-3a=0 ...(ii)
On solving Eqs. (i) and (ii), we get the point of intersection
i.e x=0,y=\(-\frac{3}{2}\)
Also, required line is parallel to X-axis.
∴ m=0
Hence, equation of line which is passing through \((0,\frac{3}{2})\) , having slope m=0 is
\(\left( y+\frac { 3 }{ 2 } \right) =0(x-0)\)
⇒ y=- \(\frac { 3 }{ 2 } \)
Thus, the required line is below X-axis at a distance of \(\frac { 3 }{ 2 } \) from X-axis.
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)
Since, a, b and c are in HP. Then, \(\frac { 1 }{ a } ,\frac { 1 }{ b } \) and \(\frac { 1 }{ c } \) are in AP
\(\therefore \frac { 2 }{ b } =\frac { 1 }{ a } +\frac { 1 }{ b } \Rightarrow \frac { 1 }{ a } -\frac { 2 }{ b } +\frac { 1 }{ c } =0\)
Hence, straight line \(\frac { X }{ a } +\frac { Y }{ b } +\frac { 1 }{ c } =0\) always passes through a fixed point (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
Let (x, y) be the coordinates of vertex C and (x1, y1) be the coordinates of centroid of the triangle.
\(\therefore \quad { x }_{ 1 }=\frac { x+2-2 }{ 3 } \)
and \({ y }_{ 1 }=\frac { y-3+1 }{ 3 } \)
⇒ \({ x }_{ 1 }=\frac { x }{ 3 } \)
and \({ y }_{ 1 }=\frac { y-2}{ 3 } \) ...(i)
Since, the centroid lies on the line 2x+ 3Y = 1.
So, point (x1, y1) satisfied the equation of line.
∴ 2x1+3y1=1
⇒ \(\frac { 2x }{ 3 } +\frac { 3(y-2) }{ 3 } =1\) [from Eq.(i)]
⇒ 2x+3y-6=3
⇒ 2x+3y=9
Hence, this is the required equation of locus of the vertex C