• # question_answer If $A\,(-2,1,)\text{ }B\,(a,0,\text{ }C\text{ (}4,b)$ and $D\text{ (}1,2)$ are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides. OR If $A\,(-5,7),\text{ }B\,(-4,-5),\text{ }C\text{ (}-1,-6)$ and $D\text{ (}4,5)$are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

 Given, ABCD is a parallelogram. Midpoint of $AC=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ $=\left( \frac{-2+4}{2},\frac{1+b}{2} \right)$ $=\left( \frac{2}{2},\frac{1+b}{2} \right)$ $=\left( 1,\frac{1+b}{2} \right)$ Midpoint of $BD=\left( \frac{x{{'}_{1}}+x{{'}_{2}}}{2},\frac{y{{'}_{1}}+y{{'}_{2}}}{2} \right)$ $=\left( \frac{a+1}{2},\frac{0+2}{2} \right)$ $=\left( \frac{a+1}{2},\frac{2}{2} \right)$ $=\left( \frac{a+1}{2},1 \right)$ Since, diagonals of a parallelogram bisect each other, $\therefore$                  $\left( 1,\frac{1+b}{2} \right)=\left( \frac{a+1}{2},1 \right)$ On comparing, we get $\therefore$      $\frac{a+1}{2}=1$                  $\frac{1+b}{2}=1$ $\Rightarrow$   $a+1=2$         $\Rightarrow$   $1+b=2$ $\Rightarrow$   $a=1$              $\Rightarrow$        $b=1$ Therefore, the coordinates of vertices of parallelogram ABCD are $A\,(-2,1),\text{ }B\,(1,0),\,\,C\,(4,1)$and $D\,(1,2)$Length of side $AB=DC=\sqrt{{{(1+2)}^{2}}+{{(0-1)}^{2}}}$ $=\sqrt{9+1}=\sqrt{10}$ units And,     $AD=BC=\sqrt{{{(1+2)}^{2}}+{{(2-1)}^{2}}}$ $=\sqrt{9+1}=\sqrt{10}$ units OR Given ABCD is quadrilateral. By joining points A and C, the quadrilateral is divided into two triangles. Now, Area of quad. ABCD = Area of $\Delta ABC$+ Area of $\Delta \,ACD$ Area of $\Delta \text{ }ABC$ $=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$ $=\frac{1}{2}\left[ -5\left( -5+6 \right)-4\left( -6-7 \right)-1\left( 7+5 \right) \right]$ $=\frac{1}{2}[-5(1)-4(-13)-1(12)]$ $=\frac{1}{2}(-5+52-12)$ $=\frac{1}{2}(35)=\frac{35}{2}$ sq. units. Area of $\Delta \text{ }ADC$ $=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$ $=\frac{1}{2}[-5(5+6)+4(-6-7)+(-1)(7-5)]$ $=\frac{1}{2}[-5(11)+4(-13)-1(2)]$ $=\frac{1}{2}(-55+52-12)$ $=\frac{1}{2}|-109|=\frac{109}{2}$ sq. units. Area of quadrilateral ABCD $=\frac{35}{2}+\frac{109}{2}$ $=\frac{144}{2}=72$ sq. units.