Tag: quadrants

Questions Related to quadrants

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.

Mid point of $A(0, 0)$ and $B(1024, 2050)$ is ${A _1}$. mid point of ${A _1}$ and B is ${A _2}$ and so on. Coordinates of ${A _{10}}$ are.

maths linear graphs quadrants the cartesian plane the cartesian system

  1. $(1022, 2044)$

  2. $(1025, 2050)$

  3. $(1023, 2046)$

  4. $(1, 2)$

Show answer
Correct Option: C

If the coordinates of the extermities of diagonal of a square are $(2,-1)$ and $(6,2)$, then the coordinates of extremities of other diagonal are 

maths linear graphs quadrants the cartesian plane the cartesian system

  1. $\left(\dfrac{5}{2},\dfrac{5}{2}\right)$

  2. $\left(\dfrac{11}{2},\dfrac{3}{2}\right)$

  3. $\left(\dfrac{11}{2},\dfrac{-3}{2}\right)$

  4. $\left(\dfrac{5}{2},-\dfrac{5}{2}\right)$

Show answer
Correct Option: C
Explanation:

$\begin{array}{l} Coordination\, \, of\, \, mid-po{ { int } }\, \, 0 \ =\left( { \frac { { 6+2 } }{ 2 } ,\frac { { 2-1 } }{ 2 }  } \right)  \ =\left( { 4,\frac { 1 }{ 2 }  } \right)  \ AB=BC \ \Rightarrow { \left( { { x _{ 1 } }-2 } \right) ^{ 2 } }+{ \left( { { y _{ 1 } }+1 } \right) ^{ 2 } }={ \left( { { x _{ 1 } }-6 } \right) ^{ 2 } }+{ \left( { { y _{ 1 } }-2 } \right) ^{ 2 } } \ \Rightarrow 8{ x _{ 1 } }+6{ y _{ 1 } }=35\to (i) \ AO=BO \ \Rightarrow { \left( { 2-4 } \right) ^{ 2 } }+{ \left( { -1-\frac { 1 }{ 2 }  } \right) ^{ 2 } }={ \left( { { x _{ 1 } }-4 } \right) ^{ 2 } }+{ \left( { { y _{ 1 } }-\frac { 1 }{ 2 }  } \right) ^{ 2 } } \ \Rightarrow 4+\frac { 9 }{ 4 } =x _{ _{ 1 } }^{ 2 }+y _{ 1 }^{ 2 }-8{ x _{ 1 } }-{ y _{ 1 } }+16+\frac { 1 }{ 4 }  \ \Rightarrow x _{ 1 }^{ 2 }+y _{ 1 }^{ 2 }-8{ x _{ 1 } }-{ y _{ 1 } }=-10\to (ii) \ from\, \, equation\, \, (i) \ put \ { x _{ 1 } }=\frac { { 35-6{ y _{ 1 } } } }{ 8 } \, \, in\, \, equation\, \, (ii) \ \Rightarrow { \left( { \frac { { 35-6{ y _{ 1 } } } }{ 8 }  } \right) ^{ 2 } }+y _{ 1 }^{ 2 }-\left( { 35-6{ y _{ 1 } } } \right) { y _{ 1 } }=-10 \ \Rightarrow 4y _{ 1 }^{ 2 }-4{ y _{ 1 } }-15=0 \ \Rightarrow \left( { 2{ y _{ 1 } }+3 } \right) \left( { { y _{ 1 } }-5 } \right) =0 \ \Rightarrow { y _{ 1 } }=\frac { { -3 } }{ 2 } ,5 \ { y _{ 1 } }=\frac { { -3 } }{ 2 } \to { x _{ 1 } }=\frac { { 35-6\times -\frac { 3 }{ 2 }  } }{ 8 } =\frac { { 11 } }{ 2 }  \ { y _{ 1 } }=5\to { x _{ 1 } }=\frac { { 35-6\times 5 } }{ 8 } =\frac { 5 }{ 8 }  \ The\, \, vertices\, of\, \, other\, \, two\, vertices\, \, are \ \left( { \frac { { 11 } }{ 2 } ,\frac { { -3 } }{ 2 }  } \right) \, \, and\, \, \left( { \frac { 5 }{ 8 } ,5 } \right)  \end{array}$