Question

. The area of a rectangle with length 6 units is 24 sq. units. The coordinates of
three of the vertices of the rectangle are (0,3,) (0,-1) and (6,3). Find the
coordinates of the fourth vertex and draw the rectangle on a Cartesian plane.
(4 mark​

Answer 1

$\huge\rm\underline\purple{GIVEN:}$

• $\large\rm{Length\:and\:area\:of\:rectangle\:is\:6units\:and\:24sq.units}$
• $\large\rm{Given\:three\:vertices\:of\:rectangle\:are\:(0,3),(0,-1),(6,3)}$

$\huge\rm\underline\purple{TO\:FIND:}$

• $\large\rm{Fourth\:vertex\:of\:rectangle}$

$\huge\rm\underline\purple{SOLUTION:}$

$\large\rm{Let\:breadth\:of\:rectangle\:be\:'b'}$

$\large\rm\bold{\boxed{Area\:of\:rectangle\:=\:L\times\:B}}$

$\large\rm{L\rightarrow{length\:of\:rectangle}}$

$\large\rm{B\rightarrow{Breadth\:of\:rectangle}}$

$\large\rm{\implies{6\times\:b\:=\:24}}$

$\large\rm{\implies{b\:=\:\frac{24}{6}\:=\:4units}}$

$\large\rm{In\:rectangle\:The\:mid\:point\:of\:diagonals\:are\:equal}$

$\large\rm{Let\:the\:fourth\:vertex\:be(x,y)}$

$\large\rm{Mid\:point\:of\:two\:points\:(x_1,y_1),(x_2,y_2)\:is}$

$\large\rm\bold{\boxed{M\:=\:(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})}}$

$\large\rm{Mid\:point\:of\:BD\:=\:(\frac{(0+6)}{2},\frac{-1+3}{2})}$

$\large\rm{Mid\:point\:of\:BD\:=\:(3,1)}$

$\large\rm{Mid\:point\:of\:AC\:=\:(\frac{x+0}{2},\frac{3+y}{2})}$

$\large\rm{Mid\:point\:of\:BD\:=\:Mid\:point\:of\:AC}$

$\large\rm{\implies{(\frac{x}{2},\frac{3+y}{2})\:=\:(3,1)}}$

$\large\rm{\implies{\frac{x}{2}\:=\:3}}$

$\large\rm{\implies{x\:=\:6}}$

$\large\rm{\implies{\frac{3-y}{2}\:=\:1}}$

$\large\rm{\implies{3+y\:=\:2}}$

$\large\rm{\implies{y\:=\:-1}}$

$\large\rm{\therefore{The\:Fourth\:vertex\:=\:(x,y)\:=\:(6,-1)}}$