Coordinate Geometry Test - 26

A median of a triangle is a line segment that joins the vertex of a triangle to the midpoint of the opposite side.

Mid point of two points $$ { (x }_{ 1 },{ y }_{ 1 }) $$ and $$ { (x }_{ 2 },{ y}_{ 2 }) $$ is  calculated by the formula $$ \left( \dfrac { { x }_{ 1 }+{x }_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 }  \right) $$

Since AD is the median, this means, D is the mid point of BC. 

Since, vertices are $$A(a, b + c),~ B(b, c + a),$$ and $$ C(c, a + b)$$
$$\therefore$$ Area $$ \bigtriangleup ABC = \dfrac{1}{2}|a(c + a) -b(b + c) + b(a + b) - c(c + a) +c(b + c) -a(a + b)|=0$$ .

The diagonals of a parallelogram bisect each other. Hence $$M$$ is the midpoint of the opposite vertices $$ (3,-2),\;(6,-3) $$ or of the opposite vertices  $$ (4,0),\;(5,-5) $$

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

2 }) $$ is  calculated by the formula $$\displaystyle \left( \frac { { x }_{ 1 }+{x}_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+y_{ 2 } }{ 2 }  \right) .$$
Using this formula, mid point of $$\displaystyle  (3, -2) , (6, -3)$$ will be,

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 $$ m:n $$, then $$(x,y) = \left( \dfrac { m{ x }_{ 2} + n{ x }_{ 1 } }{ m + n } ,\dfrac { m{ y }_{ 2 }  + n{ y }_{ 1 } }{ m + n } \right) $$

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

2 }) = (-5,7) $$  and $$ m = 2, n = 1 $$ in the section formula, we get the point 

$$ \left( \dfrac { 2(-5)  + 1(-2) }{ 2 +1 } ,\dfrac { 2(7) + 1(-2) }{ 2 + 1 } \right) =\left( -4,4 \right) $$

Midpoint of two points $$ { (x }_{ 1 },{ y }_{ 1 }) $$ and $$ { (x }_{ 2 },{ y }_{
2 }) $$ is calculated by the formula $$ \left( \cfrac { { x }_{ 1 }+{ x}_{ 2 } }{ 2 } ,\cfrac { { y }_{ 1 }+y_{ 2 } }{ 2 }  \right) $$
Using this formula, mid - point of the line-segment joining the points $$(5,4)$$ and $$(9,8)$$ is: $$= \left( \cfrac { -5 + 9 }{ 2 } ,\cfrac { 4- 8 }{ 2 }  \right)  = (2,-2) $$

Let $$ A(2,2), B(4,4) $$ and $$ C(2,6) $$ be the vertices of a given triangle ABC.

Let D, E, and F be the midpoints of AB, BC and CA respectively.

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) $$

Using this formula,the coordinates of D, E, and F are given as $$\displaystyle  D \left (\frac{2+4}{2},\frac {2+4}{2} \right ), E\left (\frac {4+2}{2},\frac {4+6}{2}\right ) $$ and $$ F\left (\dfrac {2+2}{2},\dfrac {2+6}{2} \right ) $$

i.e., $$ D(3,3), E(3,5) and F(2,4)  $$

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

}_{ 2 })$$  and $$({ x }_{ 3 },{ y }_{ 3 })$$  is $$ \left| \dfrac { {

x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 }  \right| $$

Let $$ABCD$$ be the parallelogram, having three consecutive vertices as $$A(1,-2),\ B(3,6)$$ and $$C(5,10)$$.

Also, let the fourth vertex be $$  D(x,y) $$

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

So, the midpoint of $$AC$$ is same as the midpoint of $$BD$$.

We also know that, the coordinates of the midpoint of the line segment joining $$(x_1,y_1) \ and\ (x_2,y_2)$$ are: 

$$P(x,y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$$

So, midpoint of $$ AC = $$ Midpoint of $$ BD $$

$$ \Rightarrow \left( \dfrac { 1+5 }{ 2 } ,\dfrac { -2+10 }{ 2 }  \right) =

\left( \dfrac { 3+x }{ 2 } ,\dfrac { 6 +y }{ 2 }  \right)$$

$$ \Rightarrow \left( \dfrac { 6 }{ 2 } ,\dfrac { 8 }{ 2 }   \right) =

\left( \dfrac { 3+x }{ 2 } ,\dfrac { 6 +y }{ 2 }  \right) $$

$$ \Rightarrow 3+x=6 ; \quad 6 + y = 8 $$

$$ \Rightarrow x = 3;\quad  y = 2 $$

$$ = (3,2) $$

Hence, the coordinates of the fourth vertex are $$(3,2)$$.

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 $$, then $$(x,y) = \left( \dfrac { m{ x }_{ 2

}+n{ x }_{ 1 } }{ m+n } ,\dfrac { m{ y }{ 2 }+n{ y }_{ 1 } }{ m+n } 

\right) $$
Substituting $$({ x }_{ 1 },{ y }_{ 1 }) = (1,2) $$ and $$({x }_{ 2 },{ y }_{

2 }) = (2,3) $$  and $$ m = 4, n = 3 $$ in the section formula, we get

$$ R = \left( \dfrac { 4(2) + 3(1) }{ 4 + 3 } ,\dfrac { 4(3) +3(2) }{ 4 + 3 } 

\right) =\left( \dfrac { 11}{7}, \dfrac {18}{7} \right) $$