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

Which statement is true?

  1. The x-axis is a vertical line
  2. The point $(-2 , 3)$ lies in the III quadrant

  3. Origin is the point of intersection of the x-axis and y-axis
  4. The point $(-3, -4)$ lies in the II quadrant
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

i) The x-axis ,a line parallel to it is called horizontal line

ii) The point (-2,3) lies in II Quadrant 
iii) Origin is the point of intersecting of x-axis and y-axis
iv) The point (-3,-4) lies in III quadrant .

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

Slope of the line $AB$ is $-\dfrac {4}{3}$. Co-ordinates of points $A$ and $B$ are $(x, -5)$ and $(-5, 3)$ respectively. What is the value of $x$

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

$\dfrac {y _{2} - y _{1}}{x _{2} - x _{1}} = \dfrac {-4}{3} =\dfrac{3+5}{-5-x}=\dfrac{-4}{3}$

$\Rightarrow 24 = 20 + 4x$
$x = 1$.

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

The coordinates of $A, B$ and $C$ are $(5, 5), (2, 1)$ and $(0, k)$ respectively. The value of $k$ that makes $\overline {AB} + \overline {BC}$ as small as possible is

  1. $3$
  2. $4\dfrac {1}{2}$
  3. $3\dfrac {6}{7}$
  4. $4\dfrac {5}{6}$
  5. $2\dfrac {1}{7}$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

The smallest possible value of $\overline {AC} + \overline {BC}$ is obtained when $C$ is the intersection of the y-axis, with the line that leads from $A$ to the mirror image the mirror being the y-axis) $B' : (-2, 1)$ of $B$. This is true because $\overline {CB'} = \overline {CB}$ and a straight line is the shortest path between two points. The line through and $B'$ given by
$y = \dfrac {5 - 1}{5 + 2} x + k = \dfrac {4}{7} x + k$.
To find $k$, we use the fact that the line goes through $A$:
$5 = \dfrac {4}{7} . 5 + k, k = 5 - \dfrac {20}{7} = \dfrac {15}{7} = 2\dfrac {1}{7}$;
$\therefore C$ has coordinates $(0, 2\dfrac {1}{7})$.

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

If the coordinates of vertices of a triangle is always rational then the triangle cannot be

  1. Scalene

  2. Isosceles

  3. Rightangle

  4. Equilateral
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The triangle cannot be equilateral if coordinates of vertices of the triangle is always rational.

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

The abscissa of two points A and B are the roots of the equation ${x^2} + 2ax - {b^2}$ and their ordinates are the root of the equation ${x^2} + 2px - {q^2}=0$. the equation of the circle with AB as diameter is 

  1. ${x^2} + {y^2} + 2ax + 2py + {b^2} + {q^2} = 0$
  2. ${x^2} + {y^2} - 2ax - 2py - {b^2} - {q^2} = 0$
  3. ${x^2} + {y^2} + 2ax + 2py - {b^2} - {q^2} = 0$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
${ x }^{ 2 }+2ax-{ b }^{ 2 }=0$

Let roots be ${ x } _{ 1 }$ & ${ x } _{ 2 }$

${ x } _{ 1 }+{ x } _{ 2 }=\dfrac{-b}{a}=-2a$

$ { x } _{ 1 }{ x } _{ 2 }=\dfrac{c}{a}=-{ b }^{ 2 }$

$ { x }^{ 2 }+2px-{ q }^{ 2 }=0$

Let roots be ${ y } _{ 1 }$ & ${y } _{ 2 }$

${ y } _{ 1 }+{ y } _{ 2 }=-2p$

$ { y} _{ 1 }{ y } _{ 2 }=-{ q }^{ 2 }$

Equation in diametric form is 
${ x }^{ 2 }+{ y }^{ 2 }-({ x } _{ 1 }+{ x } _{ 2 })x-({ y } _{ 1 }+{ y } _{ 2 })y+{ x } _{ 1 }{ x } _{ 2 }+{ y } _{ 1 }{ y } _{ 2 }=0$

$ { x }^{ 2 }+{ y }^{ 2 }+2ax+2py-{ b }^{ 2 }-{ q }^{ 2 }=0$