Coordinate Geometry Test - 27

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 });$$ $$({ x }_{ 2 },y_2)$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$

Let the vertices of the parallelogram be $$A (-2,-1), B(1,0),  C(4,3),  D(1,2) $$

We know that, the diagonals of a parallelogram bisect each other.

Using the section formula, if a point $$(x,y)$$ divides the line joining the

points $$({ x }_{ 1 },{ y }_{ 1 })$$ and $$({ x }_{ 2 },{ y }_{ 2 })$$ in the

ratio $$ m:n $$ internally, then $$(x,y) \equiv\displaystyle \left( \frac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m

+ n } ,\frac { m{ y }_{ 2 }  + n{ y }_{ 1 } }{ m + n }  \right) $$

Substituting $$({ x }_{ 1 },{ y }_{ 1 }) = (1,-3) $$ and $$({x }_{ 2 },{ y }_{

2 }) = (-3,9) $$  and $$ m = 1, n = 3 $$ in the section formula, we get

the point $$\displaystyle \left( \frac { 1(-3)  + 3(1) }{ 1 + 3} ,\frac { 1(9) + 3(-3)

}{1 + 3}  \right) =\left( 0,0 \right) $$

Mid-point of $$(3,4)$$ and $$(6,5)$$ is $$\left( \cfrac { 9 }{ 2 } ,\cfrac { 9 }{ 2}\right) $$
This is also mid point of diagonal joining third vertex $$(2,1)$$ and fourth vertex $$(x,y)$$
Then,
$$\cfrac { 9 }{ 2 } =\cfrac { x+2 }{ 2 } ,\cfrac { 9 }{ 2 } =\cfrac { y+1 }{ 2 } \\ \Rightarrow x=7,y=8$$

We know that, the centre of the circle lies at the midpoint of the diameter.

Also, the midpoint of line segment joining two points $$ { (x }_{ 1 },{ y }_{ 1 }) $$ and $$ { (x }_{ 2 },{ y }_{

2 }) $$ is calculated using the formula $$ \left( \dfrac { { x }_{ 1 }+{ x

}_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 }  \right) $$

Mid point of two points $$ { (x }_{ 1 },{ y }_{ 1 }) $$ and $$ { (x }_{ 2 },{ y }_{

2 }) $$ is calculated by the formula $$ \left( \dfrac { { x }_{ 1 }+{ x

}_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 }  \right) $$

Let the other point be $$ (x,y) $$

Midpoint of two points $$ { (x }_{ 1 },{ y }_{ 1 }) $$ and $$ { (x }_{ 2 },{ y }_{

2 }) $$ is  calculated by the formula $$ \left( \dfrac { { x }_{ 1 }+{ x

}_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 }  \right) $$

Using this formula, mid point of AB $$= \left( \dfrac { -5 + 7 }{ 2 } ,\dfrac { 11 + 3 }{ 2 } 

\right)$$

$$ = (1, 7) $$

Let the given vertices be $$ A (1,-12)  B (5, -10)  C(a,b) ; D(x,y) $$

We know that the diagonals of a parallelogram bisect each other. So,the midpoint of AC $$ = (2,-5) $$ and of BD $$ = (2,-5) $$. 

Let the third vertex be $$(x,x+3)$$.

Let the points be $$ A(2,-2), B(-2,1), C(5,2) $$.

Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1} \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$

Hence, Length of side AB $$ = \sqrt { \left(-2-2 \right) ^{ 2 }+\left(1 + 2\right) ^{ 2 } } = \sqrt { 16+ 9 } = \sqrt { 25 }  = 5 $$ 

Length of side BC $$ = \sqrt {\left(5 + 2\right) ^{ 2 }+\left(2-1\right) ^{ 2 } } = \sqrt { 49 + 1 } = \sqrt { 50} $$

Length of side AC $$ = \sqrt { \left(5-2 \right) ^{ 2 }+\left(2+ 2\right) ^{ 2 } } = \sqrt { 9+16 } = \sqrt { 25 }  = 5 $$

Since, $$ {(\sqrt { 50 }) }^{2} = { 5 }^{2} + { 5 }^{2} $$,
$$ \Rightarrow {BC}^{2} ={AB}^{2} + {AC}^{2} $$
Hence, the triangle has a right angle at $$A$$, with $$AB$$ and $$AC$$ as base and height.
So, area of the triangle $$ABC = \cfrac {1}{2} \times base \times height = \cfrac {1}{2} \times 5 \times 5= \cfrac {25}{2}$$ sq  units.