Coordinate Geometry Test - 29

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) $$

So, mid point of $$ (-7,14) ; (k,4) = (a,b) $$
$$ =>  \left( \dfrac { -7+k}{ 2 } ,\dfrac { 14+4

}{ 2 }  \right) \quad =\quad (a,b) $$

$$ => a = \dfrac { -7+k}{ 2 } ; b = 9 $$

Given, $$ 2a + 3b = 5 $$
$$ 2 \times \dfrac { (-7+k)}{ 2 } + 3(9) = 5 $$
$$ -7 + k + 27 = 5 $$
$$ k = -15 $$

Let the fourth vertex D $$ = (x,y) $$


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

midpoint of AC is same as the mid point of BD.


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

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

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

So, midpoint of $$ AC = $$ Mid point of $$ BD $$


$$ => \left( \frac { -2+8 }{ 2 } ,\frac { 3+3 }{ 2 }  \right) \quad =

\left( \frac { 6+x }{ 2 } ,\frac { 7 +y }{ 2 }  \right) \quad $$


$$ => \left( \frac { 6 }{ 2 } ,\frac { 6 }{ 2 }   \right) \quad =

\left( \frac { 6+x }{ 2 } ,\frac { 7 +y }{ 2 }  \right) \quad $$


$$ => 6+x=6 ; 7 + y = 6 $$


$$ => x = 0 ; y = -1 $$

Hence, $$ D = (0,-1) $$

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

Given, mid point of $$ Q(-6,5) ; R (-2,3) = P\left(\dfrac{a}{3},4\right) $$
$$\Rightarrow \left( \dfrac { -6-2 }{ 2 } ,\dfrac { 5+3 }{ 2 }  \right) =\left(\dfrac{a}{3},4\right) $$
$$\Rightarrow \dfrac{-8}{2} = \dfrac{a}{3} $$
$$\Rightarrow -24 = 2a $$
$$\Rightarrow  a = -12 $$

A

median of a triangle is a line segment that joins the vertex of a triangle to

the midpoint of the opposite side.

Median through A passed through mid point of BC.

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

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

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

Using this formula,


mid point of BC $$= \left( \frac { 1+5 }{ 2 } ,\frac { -1+1 }{ 2 } 

\right) \quad =\quad (3,0) $$

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 } } $$

Length of median through A $$ = \sqrt { \left( 3 + 1

\right) ^{ 2 }+\left(0-3 \right) ^{ 2 } } = \sqrt { 16 + 9 } = \sqrt { 25 } = 5  units

$$

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( \frac { { x }_{ 1 }+{

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

Using this formula,
mid point of AC $$= \left( \frac { 1+5 }{ 2 } ,\frac { 3+1 }{ 2 } 

\right) \quad =\quad (2,2) $$

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 median from B $$ = \sqrt { \left( 2-1

\right) ^{ 2 }+\left( 2 + 1 \right) ^{ 2 } } = \sqrt { 1 + 9 } = \sqrt {10 }

$$

The extremities of the diagonal are nothing but the opposite vertices of a parallelogram.
Let the four vertices be $$ A( 3, -4) , B (-2, 1), C (-6, 5) and D(x,y) $$

We know that the diagonals of a parallelogram bisect each
other. So,the midpoint of AC is same as the mid point of BD.
Mid point of two points $$ { (x }_{ 1 },{ y }_{ 1 }) $$ and $$ { (x }_{ 2 },{ y
}_{ 2 }) $$ is  calculated by the formula $$ \left( \frac { { x }_{ 1 }+{
x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+y_{ 2 } }{ 2 }  \right) $$
So, midpoint of $$ AC = $$ Mid point of $$ BD $$
$$ => \left( \frac { 3-6 }{ 2 } ,\frac { -4+5 }{ 2 }  \right) \quad =
\left( \frac { -2+x }{ 2 } ,\frac { 1 +y }{ 2 }  \right) \quad $$
$$ => \left( \frac { -3 }{ 2 } ,\frac { 1 }{ 2 }   \right) \quad =
\left( \frac { -2+x }{ 2 } ,\frac { 1 +y }{ 2 }  \right) \quad $$
$$ => -2 + x = -3 ; 1 + y = 1 $$
$$ => x = -1 ; y = 0 $$
Hence, $$ D = (-1,0) $$

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

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

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

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

Hence, area of the triangle with given vertices $$ \left| \frac { k(6-1)-2(1-2k)+3(2k-6) }{ 2 }  \right|  = 20 $$

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

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

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

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


Substituting $$({ x }_{ 1 },{ y }_{ 1 }) = (7,-6) $$ and $$({x }_{ 2 },{ y }_{

2 }) = (3,4) $$  and $$ m = 1, n = 2 $$ in the section formula, we get

the point $$ \left( \frac { 1(3)  + 2(7) }{ 1 +2 } ,\frac { 1(4) + 2(-6)

}{1 + 2 }  \right) = \left( \frac {17}{3}, \frac{-8}{3} \right) $$

Since, $$ x - $$ cordinate is positive and $$ y-$$ cordinate is negative, the point lies in the $$ IV $$ quadrant.

Let the coordinates of $$ B$$ be $$(a,b) $$
We know that,
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) $$

Hence, midpoint of $$ AB =  \left( \dfrac {-3 + a }{ 2 } ,\dfrac { 2 + b }{ 2 }  \right)
= (1,-2) $$

$$\Rightarrow \dfrac {-3 + a }{ 2 } = 1 ; \quad \dfrac { 2 + b }{ 2 } = -2 $$

$$\Rightarrow -3 + a = 2 ; \quad 2 + b = -4 $$
$$\therefore \  a = 5 ; \quad b = -6 $$

$$\Rightarrow \dfrac{1}{2}\bigg | a(6-1)+(-2)(1-2a)+3(2a-6) \bigg |=10$$