(18,-11)

Let P(x,y) be the point 

\(x=\frac { { m }_{ 1 }{ x }_{ 2 }+{ m }_{ 2 }{ x }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { 2\times 4+3\times (-1) }{ 2+3 } =\frac { 8-3 }{ 5 } =1\)
\(y=\frac { { m }_{ 1 }{ y }_{ 2 }+{ m }_{ 2 }{ y }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { 2(-3)+3(7) }{ 2+3 } =\frac { -6+21 }{ 5 } =\frac { 15 }{ 5 } =3\)
Then, the coordinates of points are (1,3)

\(AB=\sqrt { ({ -a-a) }^{ 2 }+({ -a-a) }^{ 2 } } =\sqrt { ({ -2a) }^{ 2 }+({ -2a) }^{ 2 } } =\sqrt { (4a^{ 2 }+4a^{ 2 } } =\sqrt { 8a^{ 2 } } =2\sqrt { 2a } \)
\(BC=\sqrt { ({ -a\sqrt { 3 } +a) }^{ 2 }+({ a\sqrt { 3 } +a) }^{ 2 } } =\sqrt { 3a^{ 2 }+a^{ 2 }-2\sqrt { 3 } a^{ 2 }+3a^{ 2 }+a^{ 2 }+2\sqrt { 3 } a^{ 2 } } =\sqrt { 8a^{ 2 } } =2\sqrt { 2a } \)
\(AC=\sqrt { ({ -a\sqrt { 3 } -a) }^{ 2 }+({ a\sqrt { 3 } -a) }^{ 2 } } =\sqrt { 3a^{ 2 }+a^{ 2 }+2\sqrt { 3 } a^{ 2 }+3a^{ 2 }+a^{ 2 }-2\sqrt { 3 } a^{ 2 } } =\sqrt { 8a^{ 2 } } =2\sqrt { 2a } \)
Since AB=BC=AC ∴△ABC is equilateral.

Since the point P(x,y) is equidistant from the points A(7, 1) and B(3, 5). 
Therefore PA = PB ...(i)
Using distance formula
\(PA=\sqrt { \left( x-7 \right) ^{ 2 }+\left( y-1 \right) ^{ 2 } } \)
\(=\sqrt { { x }^{ 2 }+{ 7 }^{ 2 }-2.7x+{ y }^{ 2 }+{ 1 }^{ 2 }-2.y.1 } =\sqrt { { x }^{ 2 }+{ y }^{ 2 }+49+1-14x-2y } \)
\(PB=\sqrt { \left( x-3 \right) ^{ 2 }+\left( y-5 \right) ^{ 2 } } \)  (using distance formula)
\(=\sqrt { { x }^{ 2 }+{ 3 }^{ 2 }-2.3x+{ y }^{ 2 }+{ 5 }^{ 2 }-2.5.y } =\sqrt { { x }^{ 2 }+9-6x+{ y }^{ 2 }+25-10y } \)
Substituting the values of PA and PB in (i). we get
\(\sqrt { { x }^{ 2 }+{ y }^{ 2 }+50-14x-2y } =\sqrt { { x }^{ 2 }+{ y }^{ 2 }-6x-10y+34 } \)
Squaring both sides, We get
\({ x }^{ 2 }+{ y }^{ 2 }+50-14x-2y={ x }^{ 2 }+{ y }^{ 2 }-6x-10y+34\)
⇒ 50-34=14x+2y-6x-10y ⇒16=8x-8y ⇒x-y=2

In \(\triangle\)ABC DE || AC,    (Given)
\(\frac { BD }{ DA } =\frac { BE }{ EC } \quad \quad \quad (BPT)\quad ...(i)\)
In \(\triangle\)ABE, DF || AE,    (Given)
\(\frac { BD }{ DA } =\frac { BF }{ FE } \quad \quad \quad (BPT)\quad ...(ii)\)
From (i) and (ii), we have
\(\frac { BF }{ FE } =\frac { BE }{ EC } \)
Hence proved.

In \(\Delta\)ACB,    LM || CB         [given]
\(\Rightarrow \quad \frac{A M}{M B}=\frac{A L}{L C}\)    ....(i)
[ by basic proportionality theorem]
In \(\Delta\)ACD, LN ||CD    [given]
\(\Rightarrow \quad \frac{A N}{N D}=\frac{A L}{L C}\)    ....(ii)
[by basic proportionality theorem]
From Eqs. (i) and (ii), we get
\(\frac{A M}{M B}=\frac{A N}{N D} \Rightarrow \frac{M B}{A M}=\frac{N D}{A N}\)
[on taking reciprocal of the terms]
\(\Rightarrow \quad \frac{M B}{A M}+1=\frac{N D}{A N}\) + 1  [adding 1 on both sides]
\(\begin{aligned} & \Rightarrow \quad \frac{M B+A M}{A M}=\frac{N D+A N}{A N} \\ \end{aligned}\)

\(\begin{aligned} \Rightarrow \quad \frac{A M}{A M+M B}=\frac{A N}{A N+N D} \end{aligned}\)
[on taking reciprocal of the terms]
\(\therefore \quad \frac{A M}{A B}=\frac{A N}{A D} \quad\left[\begin{array}{c} \because A D=A N+N D \\ \text { and } A B=A M+M B \end{array}\right]\)
Hence proved.

Let P(x, y) be the required point.
Then, P divides AB internally in the ratio 3: 2.

Here, \(\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\frac { 3 }{ 2 } \) and \(\left( { x }_{ 1 },\quad { y }_{ 1 } \right) =(4,\quad -3);\quad \left( { x }_{ 2 },\quad { y }_{ 2 } \right) =(9,\quad 7)\)
Then, \(P(x,\quad y)=P\left( \frac { { m }_{ 1 }{ x }_{ 2 }+{ m }_{ 2 }{ x }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } ,\frac { { m }_{ 1 }{ y }_{ 2 }+{ m }_{ 2 }{ y }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) \)
\( =P\left( \frac { 3\times 9+2\times 4 }{ 3+2 } ,\frac { 3\times 7+2\times (-3) }{ 3+2 } \right) \\ \ =P\left( \frac { 27+8 }{ 5 } ,\frac { 21-6 }{ 5 } \right) =P\left( \frac { 35 }{ 5 } ,\frac { 15 }{ 5 } \right) \\=P(7, 3)\)
Therefore, (7, 3) is the required point.

AB =\(\sqrt { { (5-1 })^{ 2 }+(3-0)^{ 2 } } =\sqrt { 16+9 } \) =5
DC = \(\sqrt { { (2+2 })^{ 2 }+(7-4)^{ 2 } } =\sqrt { 16+9 } \) =5
BC = \(\sqrt { { (5-2 })^{ 2 }+(3-7)^{ 2 } } =\sqrt { 9+16 } \) = 5
AD = \(\sqrt { { (1+2 })^{ 2 }+(0-4)^{ 2 } } =\sqrt { 9+16 } \) = 5
Mid - point of AC = \((\frac {1+2}{2} , \frac {0+7}{2})\) = \((\frac {3}{2} , \frac {7}{2})\); Mid-point BD = \((\frac{5-2}{2},\frac{3+4}{2})\) = \(( \frac {3}{2}, \frac {7}{2})\)
Since AB = DC and BC = AD.
Opposite sides are parallel and diagonals bisect each other.
∴ The given points are the vertices of a parallelogram.  

(i) Given, AP = 3 cm, AR = 4.5 cm, AQ = 6 cm,AB = 5 cm and AC =  10 cm
Here, \(\frac{A P}{A B}=\frac{3}{5} \text { and } \frac{A Q}{A C}=\frac{6}{10}=\frac{3}{5} \Rightarrow \frac{A P}{A B}=\frac{A Q}{A C}\)
Thus, PQ II BC [by converse of basic proportionality theorem]
In ΔARQ and ΔADC,
∠RAQ = ∠DAC [common angle]
∠ARQ = ∠ADC [corresponding angles]
∠RQA = ∠DCA [corresponding angles]
SO, ΔARQ - ΔADC [by AAA similarity criterion]
\(\Rightarrow \frac{A R}{A D}=\frac{A Q}{A C}\) [since, corresponding sides of similar triangles are proportional]
\(\Rightarrow \frac{4.5}{A D}=\frac{6}{10}\) [∵ AR = 4.5 cm , AQ = 6 cm and AC = 10 cm]
\(\Rightarrow A D=\frac{45}{6}=\frac{15}{2}=7.5 \mathrm{~cm}\)
(ii) Now, \(\frac{\operatorname{ar}(\Delta A R Q)}{\operatorname{ar}(\Delta A D C)}=\left(\frac{A Q}{A C}\right)^{2}\) [by theorem of area of similar triangles]
\(=\left(\frac{6}{10}\right)^{2}=\frac{36}{100}=\frac{9}{25}\)

Since the given points are collinear, the area of the triangle formed by them must be 0, i.e.,
\(\frac{1}{2}[2(k+3)+4(-3-3)+6(3-k)]=0\) 
i.e., \(\frac{1}{2}(-4 k)=0\) 
Therefore, k=0
Let us verify our answer.
area of \(\Delta \mathrm{ABC}=\frac{1}{2}[2(0+3)+4(-3-3)+6(3-0)]=0\) 

Join BX
In ΔABX, U is mid-Point of AB and Y is mid-Point AX
(given)
∴ UY || BX(using mid-Point theorem) ..(i)

In ΔBCX, V is mid-Point of BC and Z is mid-Point of XC
∴ VZ || BX   ....(ii)
From (i) and (ii)
UY || VZ
In ΔABC, U is mid-point of AB and V is mid-point of BE
∴ UV || AC 
⇒ UV || YZ
Hence Proved

From the given figure, it is clear that DOB is a straight line.
\(\therefore \quad \angle DOC+\angle COB={ 180 }^{ ° }\)
[by linear pair axiom]
\(\Rightarrow\quad \angle DOC+{ 125 }^{ ° }={ 180 }^{ ° }\)
[given, \(\angle COB = { 180 }^{ ° }\)]
\(\Rightarrow\quad \angle DOC={ 180 }^{ ° } - { 125 }^{ ° }={ 55 }^{ ° }\)
In \(\triangle DOC,\angle DCO+\angle CDO+\angle DOC={ 180 }^{ ° }\)
[by angle sum property of a triangle]
\(\Rightarrow\quad \angle DCO+{ 70 }^{ ° } + { 55 }^{ ° }={ 180 }^{ ° }\)
\(\because\angle CDO={ 70 }^{ ° }\) and \(\angle DOC={55 }^{ ° }]\)
\(\Rightarrow\quad \angle DCO+{ 125 }^{ ° }={ 180 }^{ ° }\)
\(\Rightarrow\quad \angle DCO={ 180 }^{ ° } - { 125 }^{ ° }={ 55 }^{ ° }\)     ... (i)
Also, \(\triangle ODC\sim \triangle OBA\)          [given]
\(\Rightarrow\quad\angle OAB=\angle OCD=\angle DCO\)
\(\angle OAB={ 55 }^{ ° }\)  [from Eq. (i)]
Hence, \(\angle DOC={ 55 }^{ ° },\angle DCO={ 55 }^{ ° }\) and \(\angle OAB={ 55 }^{ ° }\)

In \(\triangle POQ\), \(DE\parallel OQ\)   [given]
\(\therefore \frac { PE }{ EQ } =\frac { PD }{ DO } \)  ... (i)
[by basic proportionality theorem]
In \(\triangle POR\), \(DF\parallel OR\)   [given]
\(\therefore \frac { PF }{ FR } =\frac { PD }{ DO } \)... (ii)
[by basic proportionality theorem]
From Eqs.(i) and (ii),we get
\(\frac { PE }{ EQ } =\frac { PF }{ FR } \)
In \(\triangle PQR\), we have  \(\frac { PE }{ EQ } =\frac { PF }{ FR } \)
\(\therefore EF\parallel QR\)
[by converse of basic proportionality theorem]

Distance of AB = Distance of AC.