Straight Lines Test -1

Explanation:

The lines formed by these lines is right angled, triangle.

Two lines are perpendicular to each other if the product  od their slopes is -1

Slope of the lines  3x - y – 4 = 0 , x + 3y – 4 = 0 are 3 and -1/3 respectively.

The product of these slopes is -1

Hence the lines3x - y – 4 = 0 , x + 3y – 4 = 0  are perpendicular to each other.

Therefore the triangle formed by these lines is a right angled triangle.

 

Explanation:

Since the angle made with x axis is zero, since tanθ is the slope. Then tan0 = 0

Hence the slpoe of the line parallel to X-axis is zero

Option 1 is the correct answer.

 

Let L be the foot of the perpendicular from the X axis. Therefore its y coocrdinate is zero

Therefore the coordiantes of the point L is (x,0)

Hence option 1 is the correct answer

 

Explanation:

Vertical lines hve undefined slopes. Hence a line which is parallel to Y-axis has undefined slopes.

Hence option 1 is correct.

 

$$ The\quad vertices\quad of\quad the\quad given\quad \Delta ABO\quad are\\ A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 0,2y \right) ,\quad B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 2x,0 \right) \quad \& \quad O\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 0,0 \right) .\\ Let\quad the\quad point\quad I\left( p,q \right) \quad be\quad at\quad equal\quad distance\quad from\quad A,\quad B\quad \& \quad O.\quad \\ Using\quad the\quad distance\quad formula\\ d=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right)  }^{ 2 }+{ \left( { y }_{ 1 }-{ y }_{ 2 } \right)  }^{ 2 } } \quad we\quad obtain\\ AI=d=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right)  }^{ 2 }+{ \left( { y }_{ 1 }-{ y }_{ 2 } \right)  }^{ 2 } } =\sqrt { { \left( 0-{ p } \right)  }^{ 2 }+{ \left( { 2y }-{ q } \right)  }^{ 2 } } ......(i)\\ Similarly\\ BI=\sqrt { { \left( 2x-{ p } \right)  }^{ 2 }+{ \left( { 0 }-{ q } \right)  }^{ 2 } } .......(ii)\quad and\\ OI=\sqrt { { \left( 0-{ p } \right)  }^{ 2 }+{ \left( { 0 }-{ q } \right)  }^{ 2 } } =\sqrt { { p }^{ 2 }+{ q }^{ 2 } } .........(iii)\\ Since\quad AI=BI\quad We\quad have\\ \sqrt { { \left( 0-{ p } \right)  }^{ 2 }+{ \left( { 2y }-{ q } \right)  }^{ 2 } } =\quad \sqrt { { \left( 2x-{ p } \right)  }^{ 2 }+{ \left( { 0 }-{ q } \right)  }^{ 2 } } \\ \Longrightarrow { y }^{ 2 }-qy={ x }^{ 2 }-px.........(iv)\\ Again\quad BI=OI\\ \therefore \quad \sqrt { { \left( 2x-{ p } \right)  }^{ 2 }+{ \left( { 0 }-{ q } \right)  }^{ 2 } } =\sqrt { { p }^{ 2 }+{ q }^{ 2 } } \\ \Longrightarrow { x }^{ 2 }-xp=0\\ \Longrightarrow x=\left( 0,p \right) .\quad We\quad reject\quad x=0\quad since\quad 0\quad is\quad the\quad abscissa\quad of\\ the\quad vertex\quad O\quad \& \quad it\quad is\quad impossible\quad for\quad x\quad to\quad be\quad =0.\\ \therefore x=p.\quad Substituting\quad this\quad value\quad of\quad x\quad in\quad (iv)\quad we\quad have\\ { y }^{ 2 }-qy={ p }^{ 2 }-{ p }^{ 2 }=0\\ \Longrightarrow y=\left( 0,q \right) .\quad We\quad reject\quad y=0\quad since\quad 0\quad is\quad the\quad ordinate\quad of\\ the\quad vertex\quad O\quad \& \quad it\quad is\quad impossible\quad for\quad y\quad to\quad be\quad =0.\\ \therefore \quad x=p,\quad y=q.\quad i.e\quad I\left( p,q \right) =\left( x,y \right) .\\ Ans-\quad Option\quad A\\ \\NOTE-\\ A\quad point,\quad which\quad is\quad equidistance\quad from\quad the\quad vertices\quad of\quad a\quad \Delta ,\\ is\quad the\quad incentre\quad of\quad the\quad \Delta .\quad So\quad it\quad will\quad lie\quad in\quad the\quad interior\quad of\quad \\ the\quad \Delta .\\ \\ \\ \\  $$

The area of $$\triangle ABC$$ with vertices $$A\equiv(x_1, y_1)$$, $$B\equiv(x_2, y_2)$$ and $$C\equiv(x_3, y_3)$$ is given as,

$$\displaystyle \tan { \theta  } =\frac { 3 }{ 4 } ,\phi =\theta +\frac { 3\pi  }{ 4 } $$
$$\displaystyle m=\tan { \phi  } =\tan { \left( \theta +\frac { 3\pi  }{ 4 }  \right)  } =\frac { \tan { \theta  } -1 }{ 1+\tan { \theta  }  } =\frac { \frac { 3 }{ 4 } -1 }{ 1+\frac { 3 }{ 4 }  } =-\frac { 1 }{ 7 } $$
Equation of $$AC$$ is $$\displaystyle y-2=-\frac { 1 }{ 7 } \left( x-0 \right) \Rightarrow x+7y-14=0$$

The figure is a triangle.
We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$

The point are $$ A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 6,10 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 4,5 \right) $$ and $$ C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 3,8 \right) $$. 

The area of the  $$ \Delta ABC$$ with vertices 

Given that the vertices of the $$\Delta ABC$$ are $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 2,9 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( a,5 \right) $$ and $$C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 5,5 \right) and \angle B={ 90 }^{ o }$$. 

The area of a triangle is given as:

Given $$A(2,4),P(3,8),Q(7,y)$$
Also given, $$AP=AQ$$