Let the ratio be k : 1. Then by the section formula, the coordinates of the point which divides AB in the ratio k : 1 are \(\left(\frac{-k+5}{k+1}, \frac{-4 k-6}{k+1}\right)\) 
This point lies on the y-axis, and we know that on the y-axis the abscissa is 0.
Therefore, \(\frac{-k+5}{k+1}=0\) 
So, k = 5
That is, the ratio is 5 : 1. Putting the value of k = 5, we get the point of intersection as \(\left(0, \frac{-13}{3}\right)\) .

Let the coordinates of A be (x,y), O(2,-3) is the mid-point of AB

\(2=\frac { x+1 }{ 2 } ,\quad \frac { y+4 }{ 2 } =-3\)
⇒ x+1=4, ⇒ y+4=-6
⇒ x=4-1, ⇒ y=-6-4
⇒ x=3, ⇒ y=-10
Hence, the coordinates of A are (3,-10).

Let points be A(4,5), B(7,6), C(4,3) and D(1,2)
\(AB=\sqrt { \left( 7-4 \right) ^{ 2 }+\left( 6-5 \right) ^{ 2 } } =\sqrt { 9+1 } =\sqrt { 10 } \)
\(BC=\sqrt { \left( 4-7 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 } } =\sqrt { 9+9 } =3\sqrt { 2 } \)
\(CD=\sqrt { \left( 1-4 \right) ^{ 2 }+\left( 2-3 \right) ^{ 2 } } =\sqrt { 9+1 } =\sqrt { 10 } \)
\(AD=\sqrt { \left( 1-4 \right) ^{ 2 }+\left( 2-5 \right) ^{ 2 } } =\sqrt { 9+9 } =3\sqrt { 2 } \)
\(AC=\sqrt { \left( 4-4 \right) ^{ 2 }+\left( 3-5 \right) ^{ 2 } } =\sqrt { 4 } =2\)
\(BD=\sqrt { \left( 1-7 \right) ^{ 2 }+\left( 2-6 \right) ^{ 2 } } =\sqrt { 36+16 } =\sqrt { 52 } =2\sqrt { 13 } \)
Here, AB=CD=, BC=AD and AC≠BD
∴  The quadrilateral ABCD is a parallelogram.

Let D, E and F are mid-point of sides BC, CA and AB respectively of △ABC.
Area of △ABC=\(\frac{1}{2}\)|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|
=\(\frac{1}{2}\)|0(1-3)+2(3-(-1))+0(-1-1)| =\(\frac{1}{2}\)ｘ8=4 sq.units.
Area of △DEF =\(\frac{1}{4}\)хarea of △ABC=\(\frac{1}{4}\)х4 =1 sq.unit.

Let D, E and F are mid-points Of the sides AB, BC and AC respectively. 
Coordinates of D = \(\frac { 0+2 }{ 2 } ,\frac { -1+1 }{ 2 } =(1,0)\)
Coordinates of E = \(\frac { 2+0 }{ 2 } ,\frac { 1+3 }{ 2 } =(1,2)\)
Coordinates of F = \(\frac { 0+0 }{ 2 } ,\frac { 3-1 }{ 2 } =(0,1)\)
Area of △DEF = \(\frac{1}{2}\)[1x(2-1)+1(1-0)+0(-1-1)]
= \(\frac{1}{2}\)[1+1+0]=\(\frac{2}{2}\)=1 units.
Area of △ABC =\(\frac{1}{2}\)[0x(1-3)+2(3+1)+0(-1-1)]
 =\(\frac{1}{2}\)[0+8+0] = 4 sq units
The ratio of ar △DEF and ar △ABC is 1:4.   

Let P divides the line segment AB in the ratio 1:2.

\(x=\frac { 1\times -4+2\times 2 }{ 1+2 } =\frac { -4+4 }{ 3 } =0\)
and \(y=\frac { 1\times \left( -6 \right) +2\left( -3 \right) }{ 3 } \)
\( =\frac { -6-6 }{ 3 } =\frac { -12 }{ 3 } =-4\)
Hence coordinates of the point P are (0,-4).
Let Q divides the segment AB in the ratio 2:1
\({ x }_{ 1 }=\frac { 2\times \left( -4 \right) +1\times 2 }{ 2+1 } \)
\( =\frac { -8+2 }{ 3 } =\frac { -6 }{ 3 } =-2\)
\({ y }_{ 1 }=\frac { 2\times \left( -6 \right) +1(-3) }{ 2+1 } \)
\(=\frac { -12-3 }{ 3 } =\frac { -15 }{ 3 } =-5\)
Hence coordinates of the point Q are (-2,-5).

Let A(x₁,y₂), B (x₂, y₂) and c(x₃,y₃) be the vertices of a triangle D (10, 5), E (8, 4) and  F(6,6) are mid-points of sides BC, CA and  AB respectively.  
Therefore  \(\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) =\left( 6,6 \right) \)
\(\Rightarrow { x }_{ 1 }+{ x }_{ 2 }=12\) ...(i)
\({ y }_{ 1 }+y_{ 2 }=12\) ...(ii)
\( \left( \frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \right) =\left( 10,5 \right) \)
\({ x }_{ 2 }+x_{ 3 }=20\)  ...(iii)
\(\ { y }_{ 2 }+y_{ 3 }=10\) ...(iv)
\(\ \left( \frac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } \right) =\left( 8,4 \right)\)
\(\Rightarrow \ { x }_{ 1 }+x_{ 3 }=16\)
\(\ { y }_{ 1 }+y_{ 3 }=8\)  ...(v)
Adding (i), (iii) and (v) we get,
\(2({ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 })=48\) ...(vi)
\(\Rightarrow { x }_{ 1 }+{ x }_{ 2 }+x_{ 3 }=24\)
Adding (ii), (iv) and (vi) we get,
\(2({ y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 })=15\) ...(vii)
From (ii), (iv), (vi) and (xi) we get,
\({ y }_{ 1 }=5,\ { y }_{ 2 }=7,\ { y }_{ 3 }=3\) ...(viii)
Coordinates of vertices are A(4,5) B(8,7) and C(12,3).

Let coordinates of R be (x,y)

\(AR=\frac { 3 }{ 4 } AB\)
But \(AR+RB=AB\quad \Rightarrow \quad \frac { 3 }{ 4 } AB+RB=AB\)
\(\Rightarrow \ RB=AB-\frac { 3 }{ 4 } AB=\frac { 4AB-3AB }{ 4 } =\frac { AB }{ 4 } \)
\(\frac { AR }{ RB } =\frac { \frac { 3 }{ 4 } AB }{ \frac { 1 }{ 4 } AB } =\frac { 3 }{ 4 } :\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \times \frac { 4 }{ 1 } =3:1\)
Thus, R divides AB in the ratio 3:1.
\(x=\frac { 3\times 0+1\times (-4) }{ 3+1 } =\frac { 0-4 }{ 4 } =\frac { -4 }{ 4 } =-1\) and \( y=\frac { 3\times 6+1\times 0 }{ 3+1 } =\frac { 18+0 }{ 4 } =\frac { 18 }{ 4 } =\frac { 9 }{ 2 } \)
Thus , coordinates of R are \(\left( -1,\frac { 9 }{ 2 } \right) \).

\(\frac { 3\times x+4\times 2 }{ 3+4 } =-1\ \Rightarrow \ \frac { 3x+8 }{ 7 } =-1\)
3x+8=-7 ⇒ 3x=-15
x=-5
∴ Coordinates of B are(-5,-2).
\(\frac { 3\times y+4\times 2y5 }{ 3+4 } =2\ \Rightarrow \ \frac { 3y+20 }{ 7 } =2\)
3y+20=14 ⇒ 3y=14-20
3y=-6 ⇒ y=-2.

Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram.
Since, ABCD is a parallelogram.
\(\therefore\) Diagonals AC and BD will bisect each other. So, the mid-point of AC and mid-point of BD will be same 

Thus mid-point of AC = Mid-point of BD
\(\begin{aligned} \Rightarrow \quad & \left(\frac{1+x}{2}, \frac{2+6}{2}\right)=\left(\frac{4+3}{2}, \frac{y+5}{2}\right) \\ \end{aligned}\)
\(\begin{aligned} & {\left[\because \text { coordinates of mid-point }=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\right] } \end{aligned}\)
On comparing the coordinate from both sides,we get
\(\frac{1+x}{2}=\frac{4+3}{2} \text { and } \frac{2+6}{2}=\frac{5+y}{2}\)
\(\Rightarrow\) 1 + x = 7 and 8 = 5 + y
\(\therefore\)   x = 6 and y = 3

Let H(x. y) is centre of circle passing through A. B and C. Since AM. BH and CH are radius of circle. 

∴ AH=BH and BH=CH
Also AH²=BH² and BH²=CH²
AH²=(x-5)²+(y+8)²
=x²+25-10x+y²+64+16y
BH=(x-2)²+(y+9)²
=x²+4-4x+y²+81+18y
CH²=(x-2)²+(y-1)²
=x²+4-4x+y²+1-2y
∴ AH²=BH² [Radii of a circle]
∴ x²+25-10x+y²+64+16y
= x²+4-4x+y²+81+18y.
⇒ -10x+4x+16y-18y=-4
⇒ -6x-2y=-4
⇒ 3x+y=2   ---(i)
Also BH²=CH²  --- (i)
∴  x²+4-4x+16y-18y=-4
= x²+4-4x+y²+1-2y
⇒ 18y+2y=1-81
⇒ 20y= -80 ⇒ y=-4
Putting value of y in (i), we get
3x+(-4) =2 ⇒3x=2+4
⇒ 3x=6 ⇒ x=2
∴  Coordinates of centre are (2,-4).

△ABC is an equilateral triangle with  AB = BC = AC. 
Let the coordinates of C be (x, y) 
⇒ AC=AB ⇒ AC²=AB²
⇒ (x-3)²+(y-4)²=(-2-3)²+(3-4)²
⇒ x²+9-6y+y²+16-8y=(-5)²+(-1)²
⇒ x²+y²-6x--8y+25=25+1
⇒ x²+y²-6x--8y=26-25
⇒ x²+y²-6x--8y=1 ---- (i)
∴  AC=BC
∴  AC²=BC²
⇒ (x-3)²+(y-4)²=(x+2)²+(y-3)²
⇒ x²+9-6x+y²+16-8y
=x²+4+4x+y²+9-6y
⇒ 25-6x-8y=13+4x-6y
⇒ 4x+6x-6y+8y=25-13
⇒ 10x+2y=12 ⇒ 5x+y=6
⇒ y=6-5x    --- (ii)
Substituting this value of y from (ii) in (i), we get
x²+(6-5x)²-6x-8(6-5x)=1
⇒ x²+36+25x²-60x-60x-48+40x=1
⇒ 26x²-26x-12-1=0
⇒ 26x²-26x-13=0
⇒ 2x²-2x-1=0
\(\Rightarrow \ x=\frac { -(-2)\pm \sqrt { ({ -2) }^{ 2 }-4\times 2\times (-1) } }{ 4 } \)
\(\ =\frac { 2\pm \sqrt { 12 } }{ 4 } =\frac { 2\pm 2\sqrt { 3 } }{ 4 } =\frac { 1\pm \sqrt { 3 } }{ 2 } \)
Putting this value of x in equation (ii ), we get
\(y=6-5\left( \frac { 1\pm \sqrt { 3 } }{ 2 } \right) =\frac { 12-5\left( 1\pm \sqrt { 3 } \right) }{ 2 } \)
\(=\frac { 12-5\pm 5\sqrt { 3 } }{ 2 } =\frac { 7\pm 5\sqrt { 3 } }{ 2 } \)
Thus, coordinates of the third vertex are
\(\left( \frac { 1\pm \sqrt { 3 } }{ 2 } ,\frac { 7\pm 5\sqrt { 3 } }{ 2 } \right) \)

Using distance formula
\(AB=\sqrt { ({ -3-2) }^{ 2 }+[8-({ -2)] }^{ 2 } } \)
\(=\sqrt { { 5 }^{ 2 }+10^{ 2 } } =\sqrt { 25+100 } \)
\(=\sqrt { 125 } =5\sqrt { 5 } units\)
\(BC=\sqrt { [{ -1-(-3)] }^{ 2 }+(4-8)^{ 2 } } \)
\(=\sqrt { { 2 }^{ 2 }+4^{ 2 } } =\sqrt { 4+16 } =\sqrt { 20 } \)
\(=2\sqrt { 5 } units\)
\(AC=\sqrt { ({ -1-2) }^{ 2 }+[4-(-2)]^{ 2 } } \)
\(=\sqrt { { 3 }^{ 2 }+6^{ 2 } } =\sqrt { 9+36 } =\sqrt { 45 } \)
\(=3\sqrt { 5 } units\)
\(\because \ 2\sqrt { 5 } +3\sqrt { 5 } =5\sqrt { 5 } \)
∴ Points are collinear. 

Let A(7,6), B(-3,4)) be the given points and P(x,0) be the required point.
Since, P is equidistant from A and B, therefore,
AP = BP ⇒  AP²=BP²
⇒ (x-7)²+(0-6)²=(x+3)²+(0-4)² ⇒ x²+49-14x+36 = x²+9+6x+16
⇒ -14x-6x = 25-85    ⇒ -20x=-60
\(\Rightarrow x=\frac { -60 }{ -20 } =3\)
∴ Required point on axis is (3,0).