Tag: quadrants

Quizzes (0) Questions (33)

Questions Related to quadrants

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

$C$ 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

  1. $4$
  2. $6$
  3. $-4$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
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, $z=\cfrac { 4b+10a }{ a+b } =\cfrac { 4\left( 2a \right) +10\left( a \right)  }{ a+2a } =\cfrac { 18 }{ 3 } =6$
Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

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?

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a property of the intercept form of a line. For a line segment divided by a point P, the relationship between the segments and the intercepts leads to alpha + beta = 1.

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

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

  1. $\displaystyle -\frac{1}{2}$
  2. $\displaystyle \frac{1}{2}$
  3. $1$
  4. $-2$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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
$\left| \begin{matrix} k & 2-2k & 1 \ 1-k & 2k & 1 \ -k & 6-2k & 1 \end{matrix} \right| =0$

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

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

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.

  1. Yes

  2. No

  3. Ambiguous

  4. Data insufficient
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

Identify the true statement.

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $(0,0)$

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

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 _______________

  1. $(3,4)$
  2. $(1,-1)$
  3. $(7,7)$
  4. $(-1,1)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 ${ X } _{ 1 }^{ ' }O{ X } _{ 1 }\leftrightarrow { Y } _{ 1 }^{ ' }O{ Y } _{ 1 }\ =(3+4,)=(7,7)$

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

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

  1. $2bc-3ad=0$
  2. $2bc+3ad=0$
  3. $3bc-2ad=0$
  4. $3bc+2ad=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
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