P divides the joining of A and B in ratio 1:4.
The coordinates of P are \((\frac {6+4}{1+4}, \frac{7+8}{1+4})\) i.e., P(2,3).
Q divides the joining of A and B in the rario 2:3.
The coordinates of Q are \((\frac{12+3}{2+3}, \frac{14+6}{1+4})\) i.e., Q(3,4).
R divides the joining of A and B in the ratio 3:2.
The coordinates of R are \((\frac{18+2}{3+2},\frac{21+4}{3+2})\), i.e., R(4,5).

∵ A(0,2) is equidistant from the points B(3,p) and C(p,5).
∴ AB = AC ⇒ AB² = AC²
⇒ (0-3)²+(2-p)²=(0-p)²+(2-5)²
⇒ 9+4-4p+p² = p²+9
⇒ 4p=4 ⇒ p=1
Point B is (3,1)
∴ AB = \(\sqrt { { (0-3) }^{ 2 }+{ (2-1) }^{ 2 } } \)
=\(\\ =\sqrt { 9+1 } =\sqrt { 10 } \) units

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 of 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 ofa parallelograin. 

No

pentagon

\(\left( -\sqrt { 3 } ,\ \sqrt { 3 } \right) \ or\ \left( \sqrt { 3 } ,\ -\sqrt { 3 } \right) \)

Yes

No

No

A is point of intersection of line x=2 and 4x+5y=20
⇒ 4 x 2 + 5y =20  ⇒ \(y={12\over 5}\)
Coordinates of A are \(\left(2, {12\over 5}\right)\)
B is point of intersection of x=2 and y=0
Coordinates of A are (2, 0)
C is point of intersection y=0 and 4x+5y=20
⇒ 4x+5x0=20 ⇒ x=5
⇒ Coordinates of C are (5, 0)
Area ΔABC
\(={1\over 2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\)
\(={1\over 2}\left|2(0-0)+2\left(0-{12\over 5}\right)+5\left({12\over 5}-0\right)\right|\)
\(={1\over 2}\left| {-24\over 5}+12\right|={1\over2}\times{36\over 5}={18\over5}=\)sq.units= 3.6sq. units

R(x,y) lies on the line segment joining the points P(a,b) and Q(b,a). Then P,Q and R lie on a line.
⇒ x(b-a)+a(a-y)+b(y-b)=0 ⇒ bx-ax+a²-ay+by-b²=0
⇒ b(x+y)-a(x+y)+(a²-b²) =0 ⇒ (b-a)(x+y)-(b-a)(b+a)=0
⇒ (b-a){(x+y)-(b+a)}=0 ⇒ (+y)=(b+a) [ Assuming a≠ b]

Now, AP:PB = 1:2
\(p={1\times1+2\times3\over 1+2}\Rightarrow={7\over 3}\)
Also AQ:QB = 2:1\(\Rightarrow q={2\times2+1\times-4\over 1+2}=0\)

P is the point of intersection of line segment AB and line 2x-y+k=0.

Such that \(\frac {AP}{AB}=\frac{1}{3} \Rightarrow 3AP =AB\)
\(\Rightarrow \) 3AP = AP+PB ⇒ 2AP = PB
\(\Rightarrow \frac{AP}{AB} = \frac{1}{2} \Rightarrow\) AP : PB = 1:2
⇒ P divides the join of A(2,1) and B(5,-8) in the ratio 1:2.
∴ Coordinates of point P are \((\frac {5+4}{1+2}, \frac {2-8}{1+2})\) i.e, P(3,-2)
As points P lies on the 2x-y+k=0
∴ 6+2+k=0 ⇒ k=-8 

AB = \(\sqrt { ({ 2-7) }^{ 2 }+{ (3-5) }^{ 2 } } =\sqrt { 25+4 } =\sqrt { 29 } \)
BC = \(\sqrt { ({ 6-2) }^{ 2 }+{ (-7-3) }^{ 2 } } =\sqrt { 16+100 } =\sqrt { 116 } \)
CA = \( \sqrt { ({ 7-6) }^{ 2 }+{ (5+7) }^{ 2 } } =\sqrt { 1+144 } =\sqrt { 145 } \)
Since AB²+BC² = 29+116 = 145 =CA².
∴ △ABC is right angled at B.
Area = \(\frac{1}{2} AB \times BC = \frac {1}{2}=\sqrt{29}.\sqrt{116}=\frac {1}{2}\sqrt{29.2}.{2}\sqrt{29}=29\)

AB²=(-2-7)²+(5-10)² = (-9)²+(-5)² =81+25 =106
BC²=(3-(-2))²+(-4-5)²=(5)²+(-9)²=25+81=106
AC²=(3-7)²+(-4-10)²=(4)^2₊(14)²=16+196=212
Since AB²+BC²=AC²
∴ ABC is a right triangle.
AB = \(\sqrt { 106 } \) and BC = \(\sqrt { 106 } \)
∵ AB = BC
∴ ABC is an isosceles right triangle.