CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 1 - MCQExams.com

Area of a triangle ABC = $$ \frac { 1 }{ 2 } \left|[ x_{ 1 }\left( y_{ 2 }-y_{ 3 }

\right) +x_{ 2 }\left( y_{ 3 }-y_{ 1 } \right) +x_{ 3 }\left( y_{ 1 }-y_{ 2 }

\right)  \right]|$$

Substituting the given coordinates, we have

=$$ \frac { 1 }{ 2 } \left[ 5 \left(7+4) \right) -\left( 4+2\right) +7\left( 2-7) \right)  \right]$$

$$ =\frac { 1 }{ 2 } \left( 5 ( 11) +4\times( -6)
+7\times( -5)\right)$$

$$ =\frac { 1 }{ 2 } \left[(55) - ( 24) - ( 35)  \right]$$

$$ =\frac { 1 }{ 2 } \times |-4| = 2$$ square units.

Consider give the vertex of a triangle are $$(-2,0),(2,3)$$ and $$(1,-3)$$.

Let, a triangle ABC, which have vertex $$A(-2,0),B(2,3)$$ and $$C(1,-3)$$.

Now, sides of triangle AB,BC and CA are,

$$AB=\sqrt{{{\left( 2+2 \right)}^{2}}+{{\left( 3-0 \right)}^{2}}}=5unit$$

$$BC=\sqrt{{{\left( 1-2 \right)}^{2}}+{{\left( -3-3 \right)}^{2}}}=\sqrt{37}unit$$

And $$CA=\sqrt{{{\left( 1+2 \right)}^{2}}+{{\left( -3-0 \right)}^{2}}}=3\sqrt{2}unit$$

Hence, $$AB\ne BC\ne CA$$

So, required triangle is scalene triangle.

$$To\quad identify-\\ whether\quad ABCD\quad is\quad a\quad (i)\quad parallelogram,\quad (ii)\quad rectangle,\quad (iii)\quad square,\quad \\ (iv)\quad scalene\quad one.\\ Solution-\\ (i)\quad We\quad know\quad that\quad in\quad a\quad parallelogrm\quad the\quad diagonals\quad bisect\quad each\quad \\ other\quad i.e\quad the\quad mid\quad points\quad of\quad AC\quad \& \quad BD\quad are\quad same.\\ Let\quad the\quad mid\quad point\quad of\quad AC\quad be\quad P(x,y).\\ Applying\quad mid\quad point\quad theorem,\\ P(x,y)=\left( \dfrac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } ,\dfrac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 }  \right) =\left( \dfrac { -2+4 }{ 2 } ,\dfrac { -1+3 }{ 2 }  \right) =(1,1).\\ Again\quad \\ let\quad the\quad mid\quad point\quad of\quad BD\quad be\quad Q(p,q).\\ Applying\quad mid\quad point\quad theorem,\\ Q(p,q)=\left( \dfrac { { x }_{ 2 }+{ x }_{ 4 } }{ 2 } ,\dfrac { { y }_{ 2 }+{ y }_{ 4 } }{ 2 }  \right) =\left( \dfrac { 1+1 }{ 2 } ,\dfrac { 0+1 }{ 2 }  \right) =(1,1).\\ \therefore \quad P(x,y)=Q(p,q)\\ \Longrightarrow the\quad mid\quad points\quad of\quad AC\quad \& \quad BD\quad are\quad same.\\ So\quad ABCD\quad is\quad a\quad parallelogram.\\ (ii)\quad If\quad the\quad diagonals\quad of\quad the\quad parallelogram\quad ABCD\quad are\quad equal\quad \\ then\quad it\quad is\quad a\quad rectangle.\\ i.e\quad AC=BD.\\ Applying\quad distance\quad formula,\\ AC=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 3 } \right)  }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 3 } \right)  }^{ 2 } } =\sqrt { { \left( -2-4 \right)  }^{ 2 }+{ \left( -1-3 \right)  }^{ 2 } } units=2\sqrt { 13 } units.\\ BD=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 4 } \right)  }^{ 2 }+{ \left( { y }_{ 2 }-y_{ 4 } \right)  }^{ 2 } } =\sqrt { { \left( 1-1 \right)  }^{ 2 }+{ \left( 0-2 \right)  }^{ 2 } } units=2units.\\ \therefore \quad AC\neq BD.\quad \\ So\quad ABCD\quad is\quad not\quad a\quad rectagle.\\ Consequently,\quad it\quad is\quad not\quad a\quad square.\\ Ans-\quad Option\quad A.\\  $$