Straight Lines Test 0

Let the given equilateral triangle be $$ABC$$ with vertices 

Let the points be $$A=\left( { x }_{ 1 },{ y }_{ 1 } \right) =(k,-1)\quad \& \quad B=\left( { x }_{ 2 },{ y }_{ 2 } \right) =(3,2)\\ $$ 

$$ Let\quad the\quad point\quad be\quad A=\left( { x }_{ 1 },{ y }_{ 1 } \right) =(p,0)\quad \left[ since\quad it\quad is\quad on\quad the\quad X-axis \right] \quad \& \quad at\quad a\quad distance\quad of\\ d=5\quad units\quad from\quad the\quad point\quad B=\left( { x }_{ 2 },{ y }_{ 2 } \right) =(5,-4).\quad \quad \quad \quad Now\\ the\quad distance\quad between\quad two\quad points\quad \left( { x }_{ 1 },{ y }_{ 1 } \right) \quad and\quad \left( { x }_{ 2 },{ y }_{ 2 } \right) \quad is\quad d=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right)  }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right)  }^{ 2 } } \\ \therefore \quad \sqrt { { \left( 5-p \right)  }^{ 2 }+{ \left( -4-0 \right)  }^{ 2 } } =5\\ \Longrightarrow { p }^{ 2 }-10p+16=0\\ \Longrightarrow \left( p-8 \right) \left( p-2 \right) =0\\ \therefore \quad p=8\quad or\quad 2\\ So\quad the\quad points\quad are\quad \left( 8,0 \right) \quad \& \left( 2,0 \right) \\ Ans-\quad Option\quad A\\  $$

$$ The\quad point\quad A\quad is\quad on\quad Y-axis.\quad \therefore \quad its\quad abscissa=0.\quad So\quad A=\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 0,5 \right) \\ and\quad B=\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( -3,1 \right) \quad \left[ given \right] \\ \therefore \quad AB=\sqrt { ({ { x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right)  }^{ 2 } } =\sqrt { { \left( -3-0 \right)  }^{ 2 }+{ \left( 1-5 \right)  }^{ 2 } } =\sqrt { 25 } =5\quad units.\\ Ans\quad Option\quad B\quad  $$

$$ Let\quad the\quad points\quad be\quad P=(p,q),\quad Q=\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 0,0 \right) and\quad R=\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 3,\sqrt { 3 }  \right) .\\ \because \quad They\quad are\quad vertices\quad of\quad an\quad equilateral\quad \Delta ,\quad \therefore \quad PQ=QR=PR\\ So\quad PQ=\sqrt { { \left( { x }_{ 1 }-{ p } \right)  }^{ 2 }+{ \left( { y }_{ 1 }-{ q } \right)  }^{ 2 } } =\sqrt { { \left( 0-{ p } \right)  }^{ 2 }+{ \left( 0-{ q } \right)  }^{ 2 } } =\sqrt { { p }^{ 2 }+{ q }^{ 2 } } ,\\ QR=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right)  }^{ 2 }+{ \left( { y }_{ 2 }-{ { y }_{ 1 } } \right)  }^{ 2 } } =\sqrt { { \left( 3-0 \right)  }^{ 2 }+{ \left( \sqrt { 3 } -{ 0 } \right)  }^{ 2 } } =2\sqrt { 3 } and\\ PR=\sqrt { { \left( { x }_{ 2 }-p \right)  }^{ 2 }+{ \left( { y }_{ 2 }-{ q } \right)  }^{ 2 } } =\sqrt { { \left( 3-p \right)  }^{ 2 }+{ \left( \sqrt { 3 } -{ q } \right)  }^{ 2 } } =\sqrt { { p }^{ 2+ }+{ q }^{ 2 }-6p-2\sqrt { 3 } q+12 } \\ \therefore \quad PQ=QR\Longrightarrow \sqrt { { p }^{ 2 }+{ q }^{ 2 } } =2\sqrt { 3 } \Longrightarrow { p }^{ 2 }+{ q }^{ 2 }=12........(i)\\ Again\quad QR=PR\Longrightarrow \sqrt { { \left( 3-p \right)  }^{ 2 }+{ \left( \sqrt { 3 } -{ q } \right)  }^{ 2 } } =2\sqrt { 3 } \Longrightarrow { p }^{ 2 }+{ q }^{ 2 }-6p-2\sqrt { 3 } q+12=12\\ { p }^{ 2 }+{ q }^{ 2 }-6p-2\sqrt { 3 } q=0............(ii)\\ Substituting\quad { p }^{ 2 }+{ q }^{ 2 }=12\quad from\quad (i)\quad we\quad get\quad 3p+\sqrt { 3 } q=6\Longrightarrow q=\frac { 6-3p }{ \sqrt { 3 }  } ........(iii)\\ Putting\quad this\quad value\quad of\quad q\quad in(i)\quad we\quad have\quad { p }^{ 2 }+{ \left( \frac { 6-3p }{ \sqrt { 3 }  }  \right)  }^{ 2 }=12\\ \Longrightarrow 12{ p }^{ 2 }-36p=0\Longrightarrow p=\left( 0,3 \right) \\ Putting\quad these\quad values\quad of\quad p\quad in\quad (iii)\quad we\quad get\quad q=\left( 2\sqrt { 3 } ,-\sqrt { 3 }  \right) \\ \therefore \quad P=\left( 0,2\sqrt { 3 }  \right) ,\left( 3,-\sqrt { 3 }  \right) \\ Ans\quad Option\quad A $$

Let the points on the circumference be $$P=\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 5,7 \right) , Q=\left( { x }_{ 2 },{ y }_{ 2 } \right) = (6,6), R=\left( { x }_{ 3 },{ y }_{ 3 } \right)$$ $$=\left( 2,-2 \right)$$  

Given $$P(-1.5, 3), Q(6, -2)$$ and $$R(-3, 4)$$
Therefore, area is given by
$$= \frac{1}{2}\times[(-1.5)(-2-4) + 6(4-3)+(-3)(3+2)]$$
$$= \frac{1}{2}\times(9 +6-15)$$
 $$= 0$$

$$ Let\  the\  points\  be\  A=\left( { x }_{ 1 },{ y }_{ 1 } \right) ,\  \& \  B=\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 8,0 \right) .\  Then\\ d=AB=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right)  }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right)  }^{ 2 } } .\\ Let\  us\  investigate\  for\  each\  option.\\ Option\  A\longrightarrow Let\  \left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 8,-7 \right) \therefore \  AB=\sqrt { { \left( 8-8 \right)  }^{ 2 }+{ \left( 0+7 \right)  }^{ 2 } } =7\rightarrow True\\ Option\  B\longrightarrow Let\  \left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 8,7 \right) \therefore \  AB=\sqrt { { \left( 8-8 \right)  }^{ 2 }+{ \left( 0-7 \right)  }^{ 2 } } =7\rightarrow True\\ Option\  C\longrightarrow Let\  \left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 1,0 \right) \therefore \  AB=\sqrt { { \left( 8-1 \right)  }^{ 2 }+{ \left( 0-0 \right)  }^{ 2 } } =7\rightarrow True\\ Option\  D\longrightarrow Let\  \left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 0,-8 \right) \therefore \  AB=\sqrt { { \left( 8-0 \right)  }^{ 2 }+{ \left( 0+8 \right)  }^{ 2 } } =8\sqrt { 2 } \rightarrow Not\  true\\ Ans-\  Option\  D\\  $$

Any straight line which makes an angle $$\alpha$$ with the positive direction of $$X-$$axis will have their slope as