Tag: maths

Questions Related to maths

$C$ is a point on the line segment joining the points $A(2,-3,4)$ and $B(8,0,10)$. If the value of $y$-coordinate of $C$ is $-2$, then the $z-$coordinate of $C$ is

maths linear graphs quadrants the cartesian plane the cartesian system

  1. $4$

  2. $6$

  3. $-4$

  4. $5$

Show answer
Correct Option: B
Explanation:
Let C divides line joining AB in ratio $a:b$
Let $C\equiv \left( x,y,z \right) \equiv \left( x,-2,\alpha  \right) $
So, $-2=\cfrac { -3(b)+0(a) }{ a+b } $
$-2a-2b=-3b$
So, $2a=b$
So, $z=\cfrac { 4b+10a }{ a+b } =\cfrac { 4\left( 2a \right) +10\left( a \right)  }{ a+2a } =\cfrac { 18 }{ 3 } =6$

The area of the triangle formed y a tangents to the curve $2xy=a^{2}$ and the coordinates axes is

maths linear graphs quadrants the cartesian plane the cartesian system

  1. $2a^{2}$

  2. $3a^{2}$

  3. $4a^{2}$

  4. $a^{2}$

Show answer
Correct Option: A

If a point P from where line drawn cuts coordinate axes at A and B(with A on x-axis and B on y-axis) satisfies $\alpha\cdot \dfrac{x^2}{PB^2}+\beta\dfrac{y^2}{PA^2}=1$, then $\alpha +\beta$ is?

maths linear graphs quadrants the cartesian plane the cartesian system

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: A

If the points $(k, 2 - 2k), (1 - k, 2k)$ and $(-k -4, 6 - 2k)$ be collinear the possible value(s) of $k$ is/are

maths linear graphs quadrants the cartesian plane the cartesian system

  1. $\displaystyle -\frac{1}{2}$

  2. $\displaystyle \frac{1}{2}$

  3. $1$

  4. $-2$

Show answer
Correct Option: B,C
Explanation:

Let the three points be,

$A=(k,2-2k)$
$B=(1-k,2k)$
$C=(-k-4,6-2k)$

If $A,B,C$ to be collinear, the area of the triangle formed by these three points must be $0$.

For these points to be collinear, we know that
$\left| \begin{matrix} k & 2-2k & 1 \ 1-k & 2k & 1 \ -k & 6-2k & 1 \end{matrix} \right| =0$

$k(2k-6+2k)-(2-2k)[(1-k)+k]+1[(1-k)(6-2k)+2k^2]=0$
$4k^2-6k-2+2k+6-8k+2k^2+2k^2=0$
$8k^2-12k+4=0$
$2k^2-3k+1=0$
$(2k-1)(k-1)=0$
$k=1,\dfrac{1}{2}$

The points (1, -1), $\displaystyle \left ( -\frac{1}{2},\frac{1}{2} \right )$ and (1, 2) are the vertices of an isosceles triangle

Say yes or no.

maths linear graphs quadrants the cartesian plane the cartesian system

  1. Yes

  2. No

  3. Ambiguous

  4. Data insufficient

Show answer
Correct Option: A
Explanation:

Let the point (1, -1), $\displaystyle \left ( -\frac{1}{2},\frac{1}{2} \right )$ and (1, 2) be denoted by P, Q and R respectively
Now PQ = $\displaystyle \sqrt{\left ( -\frac{1}{2}-1 \right )^{2}+\left ( \frac{1}{2}+1 \right )^{2}}=\sqrt{\frac{18}{4}}=\frac{3}{2}\sqrt{2}$
QR = $\displaystyle \sqrt{\left ( 1+\frac{1}{2} \right )^{2}+\left ( 2-\frac{1}{2} \right )^{2}}=\sqrt{\frac{18}{4}}=\frac{3}{2}\sqrt{2}$
PR = $\displaystyle \sqrt{\left ( 1-1 \right )^{2}+\left ( 2+1 \right )^{2}}=\sqrt{9}=3 $
From the above we see that PQ = QR
$\displaystyle \therefore $ The triangle is isosceles

The point $(5,\,3)$ lies on line $3x+2y=18$

maths linear graphs quadrants the cartesian plane the cartesian system

  1. Yes

  2. No

  3. Cant predict

  4. None

Show answer
Correct Option: B
Explanation:

Put $x=5,\,y=3$ in the given equation
LHS: $3x+2y=3\times5+2\times3$
$=15+6=21$
LHS $\neq RHS$
It does not lie on line $3x+2y=18$

Identify the true statement.

maths linear graphs quadrants the cartesian plane the cartesian system

  1. The $X$-axis is a vertical line

  2. The $Y$-axis is a horizontal line

  3. The scale on both the axes must be the same in a Cartesian plane

  4. The point of intersection between the $X$-axis and $Y$-axis is called the origin

Show answer
Correct Option: D
Explanation:

1) A line parallel to x-axis is called a horizontal line parallel to y-axis is called a vertical line.

2) The point of intersection between the X-axis and Y-axis is called the origin $(0,0)$

If a point $P$ has coordinates $(3,4)$ in a coordinate system $X'OX\leftrightarrow  Y'OY$, and if $O$ has coordinates $(4,3)$ in another system ${X} {1}'{O} _{1}{X} _{1}\leftrightarrow  {Y} _{1}'{O} _{1}{Y} _{1}$ with $X'OX\parallel  {X} _{1}'{O} _{1}{X} _{1}$, then the coordinates of $P$ in the new system ${X} _{1}'{O} _{1}{X} _{1}\leftrightarrow  {Y} _{1}'{O} _{1}{Y} _{1}$ is _______________

maths linear graphs quadrants the cartesian plane the cartesian system

  1. $(3,4)$

  2. $(1,-1)$

  3. $(7,7)$

  4. $(-1,1)$

Show answer
Correct Option: C
Explanation:

Given the point P has coordinate (3,4).Similarly O has (4,3)

Given that $X'OX\leftrightarrow Y'OY\ { X } _{ 1 }^{ ' }{ O } _{ 1 }{ X } _{ 1 }\leftrightarrow { Y } _{ 1 }^{ ' }{ O } _{ 1 }{ Y } _{ 1 }\ \Longrightarrow { X }^{ ' }OX\parallel { X } _{ 1 }^{ ' }{ O } _{ 1 }{ X } _{ 1 }$
$\therefore$ Coordinate of P in new system ${ X } _{ 1 }^{ ' }O{ X } _{ 1 }\leftrightarrow { Y } _{ 1 }^{ ' }O{ Y } _{ 1 }\ =(3+4,)=(7,7)$

Let a, b, c and d be non-zero numbers. If the point of intersection of the lines $4ax+2ay+c=0$ and $5bx+2by+d=0$ lies in the fourth quadrant and is equidistant from the two axes then

maths linear graphs quadrants the cartesian plane the cartesian system

  1. $2bc-3ad=0$

  2. $2bc+3ad=0$

  3. $3bc-2ad=0$

  4. $3bc+2ad=0$

Show answer
Correct Option: C
Explanation:
Solve the equations of two lines to get a point and satisfy it in the third equation of a line.

$4ax+2ay+c=0$ and $5bx+2by+d=0$

Let the point of intersection be $(k, -k)$

From both equations, we get 

$4ak-2ak+c=0\ ,\ 5bk-2bk+d=0$

$2ak+c=0\ ,\ 3bk+d=0$

$k=\dfrac{-c}{2a}\ ,\ k=\dfrac{-d}{3b}$

$\dfrac{-c}{2a}=\dfrac{-d}{3b}$

$\therefore 3bc-2ad=0$

Option C

The points $A\left( {2a,\,4a} \right),\,B\left( {2a,\,6a} \right)\,$ and $C\left( {2a + \sqrt 3 a,\,5a} \right)$ (when $a>0$) are vertices of 

maths linear graphs quadrants the cartesian plane the cartesian system

  1. an obtuse angled triangle

  2. an equilateral triangle

  3. an isosceles obtuse angled triangle

  4. a right angled triangle

Show answer
Correct Option: B
Explanation:

Consider the given point

$A\left( 2a,4a \right),\,B\left( 2a,6a \right)\,and\,C\left( 2a+\sqrt{3}a,5a \right)$

Distance between AB,BC and CA  respectively.

$ AB=\sqrt{{{\left( 2a-2a \right)}^{2}}+{{\left( 4a-6a \right)}^{2}}} $

$ =\sqrt{0+{{(-2a)}^{2}}} $

$ =\sqrt{4{{a}^{2}}}=2a $

$ BC=\sqrt{{{\left( 2a-(2a+\sqrt{3}a) \right)}^{2}}+{{\left( 6a-5a \right)}^{2}}} $

$ =\sqrt{3{{a}^{2}}+{{a}^{2}}} $

$ =\sqrt{4{{a}^{2}}}=2a $

$ CA=\sqrt{{{\left( 2a+\sqrt{3}a-2a \right)}^{2}}+{{\left( 5a-4a \right)}^{2}}} $

$ =\sqrt{3{{a}^{2}}+{{a}^{2}}} $

$ =\sqrt{4{{a}^{2}}}=2a $

Hence, $AB=BC=CA$

It is an equilateral triangle.

Option (B) is correct answer.