$\text{Any point on the y-axis has coordinates}\text{of the form}(0,y).\text{Since the point is equidistant from both}\text{points}(-3,4)\text{and}(7,6)\text{by distance formula,}{3}^2+{(y-4)}^2=7^2+{(y-6)}^{2}16–8y=40+36–12y4y=60y=15\therefore \text{the point is}(0,15).$

Area of a triangle with vertices $(x_1,y_1),(x_2,y_2)and(x_3,y_3)$ is given by
$Area=\frac{1}{2}\times |\left(x_1\right)(y_2-y_3)+\left(x_2\right)(y_3-y_1)+\left(x_3\right)(y_1-y_2)|$
Thus, $\text{Area of the given triangle}$
$=\frac{1}{2}|a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)|$
$=\frac{1}{2}|ac-ab+ab-bc+bc-ac|=0$

From section formula, if a point $(x,y)$ divides the line segment joining the points ($x_1,y_1$) and ($x_2,y_2$) internally in the ratio m:n,  then $x=\frac{m{x}_2+n{x}_1}{m+n}$ and $y=\frac{m{y}_2+n{y}_1}{m+n}$.

So, $2=\frac{4m+n(-1)}{m+n}$

$⟹4m-n=2m+2n$

i.e., $2m=3n$

Thus,  $\frac{m}{n}$ = $\frac{3}{2}$  or $m:n=3:2$

The mid-point of the coordinates $(x_1,y_1),(x_2,y_2)$ is given by $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
$\therefore$ The mid-points of AB, BC and CA are $(\frac{-4+2}{2}$,  $\frac{3+3}{2})$, $(\frac{4+2}{2}$,  $\frac{3+5}{2})$ and $(\frac{-4+4}{2}$,  $\frac{5+3}{2})$ respectively. Let us denote these by D, E and F.
i.e.,  D(-1, 3), E(3, 4) and F(0, 4).

Area of the triangle formed by $(x_1,y_1),(x_2,y_2)(x_3,y_3)$ is given by
$\text{Area}=\frac{1}{2}$|$x_1$($y_2$-$y_3$)+$x_2$($y_3$-$y_1$)+$x_3$($y_1$-$y_2$)|
Now, let $x_1$ = -1,  $y_1$ = 3, $x_2$ = 3,  $y_2$ = 4 $x_3$ = 0 and  $y_3$ = 4

$\Rightarrow \text{Required area}=\frac{1}{2}$| -1(4 - 4) + 3(4 - 3) + 0(3 - 4)|  = 1.5 square units

$\therefore$ Area of the triangle = 1.5 square units

Consider the points 
$A(-4,-2),B(-3,-5),C(3,-2)$ and $D(2,3).A\left(△\right)=\frac{1}{2}\times |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$\therefore$ Area of  $△ABC$ 

$=\frac{1}{2}|(-4)(-5+2)-3(-2+2)+3(-2+5)|=\frac{1}{2}|20-8-6+15|=\frac{21}{2}A\left(△ABC\right)=10.5\text{sq. units}$

Similarly, area of   $△ACD$

$=\frac{1}{2}|(-4)(-2-3)+3(3+2)+2(-2+2)|=\frac{1}{2}|20+15|=\frac{35}{2}A\left(△ACD\right)=17.5\text{sq. units}$

$\text{Area of quadrilateral}ABCD=A\left(△ABC\right)+A\left(△ACD\right)=(10.5+17.5)=28\text{sq. units}$

We have to find a point on the y-axis. Therefore, its x-coordinate will be 0.
Let the point on y-axis be P (0,y)
Distance PA = Distance between $(0,y)$ and $(3,-4)=\sqrt{(0-3{)}^2+(y-(-4){)}^2}=\sqrt{(-3{)}^2+(y+4{)}^2}$
Distance PB = Distance between $(0,y)$ and $(-5,9)=\sqrt{(0-(-5){)}^2+(y-9{)}^2}=\sqrt{5^2+(y-9{)}^2}$
By the given condition, PA = PB, thus
$\sqrt{(-3{)}^2+(y+4{)}^2}=\sqrt{5^2+(y-9{)}^2}$
Squaring on both sides of the equation
$\therefore$ $(y+4{)}^2+9=(y-9{)}^2+25$
$y^2+16+8y+9=y^2-18y+81+25$
$26y=106-25$
$26y=81$
$y=\frac{81}{26}$
$\therefore$ The required point is $(0,\frac{81}{26})$.

If a point P(x,y) divides the line joining $(x_1,y_1)\text{and}(x_2,y_2)$ in the ratio m:n
then $x$ = $\frac{m{x}_2+n{x}_1}{m+n}$ and y = $\frac{m{y}_2+n{y}_1}{m+n}$
$\therefore x$ = $\frac{m{x}_2+n{x}_1}{m+n}$ = $\frac{2(-5)+1(-2)}{2+1}$ = -4

and $y$ = $\frac{m{y}_2+n{y}_1}{m+n}$ = $\frac{2\left(7\right)+1(-2)}{2+1}$ = 4
$\therefore$ the required point is (-4,4).

The distance of a point $(x,y)$ from the origin $(0,0)$ is $\sqrt{(x-0{)}^2+(y-0{)}^2}=\sqrt{x^2+y^2}$

Let $C(x,0)$ divide $AB$ in the ratio $k:1$.

By section formula, the coordinates of $C$ are given by: $(\frac{6k+1}{k+1},\frac{4k-7}{k+1})$

But $C(x,0)=(\frac{6k+1}{k+1},\frac{4k-7}{k+1})$ $\Rightarrow \frac{4k-7}{k+1}=0$

$\Rightarrow 4k-7$ = 0 $\Rightarrow k$ = $\frac{7}{4}$

i.e., the x-axis divides $AB$ in the ratio $7:4$.

Area of a triangle formed by $(x_1,y_1),(x_2,y_2)(x_3,y_3)$ = $\frac{1}{2}$|$x_1$($y_2$ - $y_3$) + $x_2$($y_3$ - $y_1$)+ $x_3$($y_1$ - $y_2$)| 
Thus, the area of  $\Delta ABC$ formed by the given points A(4, 1), B(-3, 2) and C(0, k) is

$=\frac{1}{2}|4(2-k)+(-3)(k-1)+0(1-2)|$

$=\frac{1}{2}|8-4k-3k+3|=\frac{1}{2}|11-7k|$

But given that the area of $\Delta ABC=12\text{sq. unit}$

 $\Rightarrow |11-7k|=24$

$\pm (11-7k)=24$            $\Rightarrow 11-7k=24$ or $-(11-7k)=24$

$If11-7k=24,then-7k=24-11=13$       $\Rightarrow k=\frac{-13}{7}$

$-(11-7k)=24⟹-11+7k=24$

  $\Rightarrow 7k=24+11=35$    $\Rightarrow k=\frac{35}{7}=5$

Hence the values of k are : 5,   $\frac{-13}{7}$