Let Q(x, 0) be any point on the x-axis which lies on the perpendicular bisector of the line segment joining the point A(-5, -2) and B(4, -2)
⇒ Q is equidistant from A and B
⇒ |AQ| = |BQ|
⇒ AQ² = BQ²
⇒ (x+5)²+(0+2)²=(x-4)²+(0+2)²
⇒ x²+10x+25+4=x²-8x+16+4
⇒ 18x=-9
⇒  \(x=-{9\over 18}\)
⇒ \(x=-{1\over 2}\)
Thus, the required point Q on the x-axis is \(\left(-{1\over 2},0\right)\).Since Q lies on the perpendicular bisector of the line segment joining A and B ⇒ |AQ|=|BQ|.
Hence, triangle formed by the points Q, A and B is an isosecles triangle.

Let the coordinates of B and C be (a,b) and (x,y) respectively
Then \(\left( \frac { 1+a }{ 2 } ,\frac { -4+b }{ 2 } \right) \)  =(2.-1)

\(\Rightarrow \) \(\frac { 1+a }{ 2 } =2\quad and\quad \frac { -4+b }{ 2 } =-1\)
\(\Rightarrow \) 1+a=4 and -4+b=-2
\(\Rightarrow \) a=3 and b=2
Also  \(\left( \frac { 1+x }{ 2 } ,\frac { -4+y }{ 2 } \right) \)=(0,-1)
\(\Rightarrow \) \(\frac { 1+x }{ 2 } =0\quad and\quad \frac { -4+y }{ 2 } =-1\)
\(\Rightarrow \)  1+x=0 and -4+y=-2
\(\Rightarrow \)   x=-1 and y=2
Thus the coordinates of the vertices of ABC are A(1,-4),B(3,2) and C(-1,2)
Area of ABC =\(\frac { 1 }{ 2 } \) |1(2-2)+3(2+4)+(-1)(-4-2)|
= \(\frac { 1 }{ 2 } \)|0+18+6|
= \(\frac { 1 }{ 2 } \)(24)=12 sq.units 

26 sq.units

k=-1

p=8

Let (– 4, 6) divide AB internally in the ratio m₁ : m₂.Using the section formula, we get
\((-4,6)=\left(\frac{3 m_{1}-6 m_{2}}{m_{1}+m_{2}}, \frac{-8 m_{1}+10 m_{2}}{m_{1}+m_{2}}\right)\) 
Recall that if (x, y) = (a, b) then x = a and y = b.
So,\(-4=\frac{3 m_{1}-6 m_{2}}{m_{1}+m_{2}}\) and \(6=\frac{-8 m_{1}+10 m_{2}}{m_{1}+m_{2}}\) 
Now, \(-4=\frac{3 m_{1}-6 m_{2}}{m_{1}+m_{2}}\) gives us
– 4m₁ – 4m₂ = 3m₁ – 6m₂
i.e., 7m₁ = 2m₂
i.e., m₁ : m₂ = 2 : 7
You should verify that the ratio satisfies the y-coordinate also.
Now,\(\frac{-8 m_{1}+10 m_{2}}{m_{1}+m_{2}}=\frac{-8 \frac{m_{1}}{m_{2}}+10}{\frac{m_{1}}{m_{2}}+1}\) ,(Dividing throughout by m₂)
\(=\frac{-8 \times \frac{2}{7}+10}{\frac{2}{7}+1}=6\) 
Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7.
Alternatively : The ratio m₁ : m₂ can also be written as \(\frac{m_{1}}{m_{2}}: 1\) ,or k : 1. Let (– 4, 6) divide AB internally in the ratio k : 1. Using the section formula, we get
\((-4,6)=\left(\frac{3 k-6}{k+1}, \frac{-8 k+10}{k+1}\right)\) 
So, \(-4=\frac{3 k-6}{k+1}\) 
i.e., – 4k – 4 = 3k – 6
i.e., 7k = 2
i.e., k : 1 = 2 : 7
You can check for the y-coordinate also.
So, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7.

\(P({13\over 2},{9\over 2})\), Q(8,5), R\(({19\over 2}, {11\over2})\)

Let P and Q be the points of trisection of AB i.e., AP = PQ = QB
Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by applying the section formula, are
\(\left(\frac{1(-7)+2(2)}{1+2}, \frac{1(4)+2(-2)}{1+2}\right), \)i.e., (–1, 0)
Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are
\(\left(\frac{2(-7)+1(2)}{2+1}, \frac{2(4)+1(-2)}{2+1}\right)\) ,i.e., (– 4, 2)
Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (– 4, 2).

(10,3), (13,1)

Given vertices of a quadrilateral ABCD are
A(-5, 7), B(-4, - 5), C(-l, - 6) and D(4, 5).

\(\therefore\) Area of quadrilateral ABCD
\(=\text { Area of } \Delta A B C+\text { Area ot } \triangle A C D\)   ...(i)
Now, area of \(\triangle A B C\) 
\(=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|\) 
\(=\frac{1}{2} \mid-5(-5+6)-4(-6-7)-1(7+5)\) 
\(=\frac{1}{2}|-5 \times 1-4(-13)-1 \times 12|\) 
\(=\frac{1}{2}|-5+52-12|=\frac{|35|}{2}=\frac{35}{2} \text { sq units }\) 
and area of \(\Delta A C D=\frac{1}{2} \mid-5(-6-5)-1(5-7)+4(7+6) \mid\) 
\(=\frac{1}{2}|-5 \times(-11)-1 \times(-2)+4 \times 13|\) 
\(=\frac{1}{2}|55+2+52|=\frac{|109|}{2}=\frac{109}{2} \mathrm{sq} \text { units }\) 
\(\therefore\) From Eq. (i) say,
Area of quadnlateral ABCD = \(\frac{35}{2}+\frac{109}{2}\) 
\(=\frac{144}{2}=72 \mathrm{sq} \text { units }\)