Coordinate Geometry Test - 22

$$ Given\quad that\quad the\quad point\quad \quad R\left( p,q \right) =\left( 2,5 \right) \quad is\quad the\quad mid\quad point\quad of\quad the\quad line\quad segment\\ PQ\quad joining\quad the\quad points\quad P=\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 0,y \right) \quad and\quad Q=\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( x,0 \right) .\\ \therefore \quad by\quad mid-point\quad formula\quad we\quad have\\ p=\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } \quad and\quad q=\frac { y_{ 1 }+y_{ 2 } }{ 2 } \quad .\\ \Longrightarrow p=\frac { 0+x }{ 2 } =\frac { x }{ 2 } \quad and\quad q=\frac { y+0 }{ 2 } =\frac { y }{ 2 } \\ \Longrightarrow \frac { x }{ 2 } =2\quad i.e\quad x=4\quad and\quad \frac { y }{ 2 } =5\quad i.e\quad y=10\\ \therefore \quad The\quad coordinates\quad of\quad P\quad is\quad \left( 0,y \right) =\left( 0,10 \right) \quad and\\ the\quad coordinates\quad of\quad Q\quad is\quad \left( x,0 \right) =\left( 4,0 \right) \\ Ans-\quad Option\quad D\\  $$

The figure is a triangle.
We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$

Let the points given be $$A = (7,6)$$ and $$B = (3,4)$$. Let the point which divides it internally in the ratio $$1:2$$ be $$P$$

The area of the  $$ \Delta ABC$$ with vertices 

Given that the vertices of the $$\Delta ABC$$ are $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 2,9 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( a,5 \right) $$ and $$C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 5,5 \right) and \angle B={ 90 }^{ o }$$. 

Given that the point $$R\left( x,y \right)$$ divides the line segment $$PQ$$ joining the points $$P\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 1,3 \right) \ \&\ Q\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 2,5 \right)$$ in a ratio such that $$\dfrac { PR }{ PQ } =\dfrac { 3 }{ 5 }$$ i.e $$PR<PQ$$.

$$\Rightarrow R$$ divides $$PQ$$ internally. 

Here $$\dfrac { PR }{ PQ } =\dfrac { 3 }{ 5 }$$ 

$$\Rightarrow \dfrac { PQ }{ PR } =\dfrac { 5 }{ 3 }$$

$$\Rightarrow \dfrac { PR+RQ }{ PR } =\dfrac { 5 }{ 3 }$$

$$\Rightarrow \dfrac { RQ }{ PR } =\dfrac { 2 }{ 3 }$$

$$\Rightarrow \dfrac { PR }{ QR } =\dfrac { 3 }{2 }$$

$$ \therefore m:n=3:2$$

$$\therefore$$ By the section formula, we have

$$x=\dfrac { n{ x }_{ 1 }+m{ x }_{ 2 } }{ m+n } \&\ y=\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n }$$

i.e $$x=\dfrac { 2\times 1+3\times 2 }{ 2+3 } =\dfrac { 8 }{ 5 }$$

and $$y=\dfrac { 2\times 3+3\times 5 }{ 2+3 } =\dfrac { 21 }{ 5 }$$

Therefore, $$x= \dfrac { 8 }{ 5 }; y=\dfrac { 21 }{ 5 }  $$

Hence, option A is the correct answer.

$$ The\quad points\quad A(x_{ 1 },y_{ 1 }),\quad B(x_{ 2 },y_{ 2 })\quad and\quad C(x_{ 3 }y_{ 3 })\quad are\quad the\quad vertices\quad of\quad \Delta ABC.\\ \therefore \quad The\quad median\quad AD\quad from\quad A\quad will\quad meet\quad BC\quad at\quad D\quad which\quad is\quad the\quad \\ mid\quad point\quad of\quad BC.\\ So,\quad by\quad section\quad formula\quad for\quad mid\quad point,\quad the\quad co-ordinates\quad of\quad D\quad is\\ D\left( \frac { x_{ 2 }+x_{ 3 } }{ 2 } ,\frac { y_{ 2 }+y_{ 3 } }{ 2 }  \right) .\\ Ans-\quad D\left( \frac { x_{ 2 }+x_{ 3 } }{ 2 } ,\frac { y_{ 2 }+y_{ 3 } }{ 2 }  \right) . $$

The point are $$ A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 6,10 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 4,5 \right) $$ and $$ C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 3,8 \right) $$. 

Let a point $$P(x,y)$$ divide the line segment $$AB$$ joining the points $$A\left( { x }_{ 1 },{ y }_{ 1 } \right)\ \&\ B\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ in the ratio $$m:n$$.