$1^{st}$ method: - In a right angled triangle

Circumradius = $\frac{hypotenuse}{2}$

The coordinates of circumcentre = (3,1)

$2^{nd}$ method: - If O(x₁, y₁) is the circumcentre, then $O{A}^2=O{B}^2$.
$(x_1-2{)}^2+(Y_1-3{)}^2=(X_1-4{)}^2+(Y_1+1{)}^2$
$-4X_1-6Y_1+8X_1-2Y_1=17-13=4$
i.e., $4X_1-8Y_1=4$ or $X_1-2Y_1=1$ ..................(I)
$O{B}^2=O{C}^2$
$X_1-4{)}^2+(Y_1+1{)}^2=(X_1-4{)}^2+(Y_1-3{)}^2$
$2Y_1+6Y_1=8$
$8Y_1=8$ ....................(II)
$Y_1=1$
From (i) $X_1=3$
$X_1,Y_1)=(3,1)$

Given lines are 5x + 12y + 18 = 0……………….(i) and 5x + 12y – 21 = 0………………….(ii)

Length of the side = distance between lines (i) and (ii) is given by

D = $|\frac{C_2-C_1}{\sqrt{a^2+b^2}}|=|\frac{-21-18}{\sqrt{25+144}}|=|\frac{-39}{13}|=3$

Length of the diagonal = $3\sqrt{2};$ Option(a).

The simplest way to find this is to put a rectangle around the given triangle and then subtract off the areas of the right triangles that surround it. (See Figure) The area of the entire rectangle is 126, but after subtracting the areas of the 3 right triangles, whose areas are 54, 3, and 42, we are left with an area of 27.

$1^{st}$ method: - The simplest way to find this is to put a rectangle around the given quadrilateral and then subtract off the areas of the right triangles that surround it. (See Figure) The area of the entire rectangle is 40, but after subtracting the areas of the 4 right triangles, whose areas are $\frac{5}{2}$, 4, 9 and 3. we are left with an area of 21.5.

$2^{nd}$ method: - Area of quadrilateral = $2(x_1{y}_2-x_2{y}_1)+(x_2{y}_3-x_3{y}_2)+(x_3{y}_4-x_4{y}_3)+(x_4{y}_1-x_1{y}_4)$
=$\frac{1}{2}(-4\times 0-(-1)\times 2)+(-1\times 1-4\times 0)+(4\times 5-2\times 1)+(2\times 2-(-4)\times 5)=21.5$

2x + 2y + 1 = 0 …………….(i)

8x – 2y + 9 = 0 …………………………(ii)

After solving equation (i) and (ii)

x = - 1 and y = $\frac{1}{2}$

$(-1,\frac{1}{2})$; which is in $2^{nd}$ quadrant.

Given, AB = 2BC

AB : BC = 2 : 1

Let C BE (h, k), then

$(\frac{2h\pm 1}{2\pm 1};\frac{2k\pm 7}{2\pm 1})$ = (6, - 3) [B may divide AC internally or externally]

2h + 1 = 18 and 2k + 7 = - 9

(or) 2h - 1 = 6 , 2k - 7 = - 3

2h = 17 and 2k = - 16 (or) 2h = 7, 2k = 4

h = $\frac{17}{2}$ and k = - 8 (or) h = $\frac{7}{2}$; k = 2

C = $(\frac{17}{2};-8)$ or $(\frac{17}{2};2)$

Option(b)

Option(a)

Slope of the line = $\frac{y_2-y_1}{x_2-x_1}=\frac{7-5}{9-7}=1$ 

Here $(x_1,y_1)$ = (7, 5) and $(x_2,y_2)$ = (9, 7)

Even without calculations it is clear that $(\sqrt{2},\frac{31}{3})$ cannot lie on the line y = 5x + 3. If x is irrational, y must also be irrational. Option(d)

Area of pentagon = Area of triangle ABE + Area of rectangle BCDE

$\frac{1}{2}\times 4\times 2+5\times 4=24$

A(1, - 2) ,B(- 3, 0) and C(5, 6)

AB = $\sqrt{20}$, BC = $\sqrt{100}$ , AC = $\sqrt{80}$

$B{C}^2=A{B}^2+A{C}^2$

ABC is a right angled triangle with the right angle at A.

Circumradius is half the length of the hypotenuse.

R = $\frac{1}{2}\times$ BC, BC = 5. Option(a).