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CAT 2025 Lesson : Trigonometry & Coordinate - Triangle, Quadrilateral & Circle

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4.2 Area Centroid and Incentre of Triangle

Where P (
x1,y1x_1, y_1), Q (x2,y2x_2, y_2), R (x3,y3x_3, y_3) are three points on a plane, then

Area of △PQR =
12×x1x2x3x1y1y2y3y1=12×(x1y2x2y1)+(x2y3x3y2)+(x3y1x1y3)\dfrac{1}{2} \times \begin{vmatrix}x_1 & x_2 & x_3 & x_1 \\ y_1 & y_2 & y_3 & y_1\end{vmatrix} = \dfrac{1}{2} \times |(x_1 y_2 - x_2 y_1) + (x_2 y_3 - x_3y_2) + (x_3 y_1 - x_1y_3)|

The area can also be written as follows (this might be easier to remember as well)

Area of △PQR =
12×x1x2y1y2x2x3y2y3=12×(x1x2)(y2y3)(x2x3)(y1y2)\dfrac{1}{2} \times \begin{vmatrix}x_1 & x_2 & y_1 & y_2 \\ x_2 & x_3 & y_2 & y_3\end{vmatrix} = \dfrac{1}{2} \times |(x_1 - x_2)(y_2 - y_3) - (x_2 - x_3)(y_1 - y_2)|

Centroid of △PQR has coordinates
(x1+x2+x33,y1+y2+y33)\left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3} \right)

Where
a,ba, b and cc are the lengths of the sides opposite the points P, Q and R,

Incentre of △PQR has coordinates
(ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c)\left(\dfrac{ax_1 + bx_2 + cx_3}{a + b + c}, \dfrac{ay_1 + by_2 + cy_3}{a + b + c} \right)

Example 17

If the coordinates of the triangle are (5, 5), (14, 5) and (5, 17)

(i) What are the coordinates of the incentre?
(ii) What are the coordinates of the centroid?
(iii) What is the area of the triangle?

Solution

(i) (a,b)(a, b) are the co-ordinates of the incentre of △PQR.

x1=5,x2=14,x3=5x_1 = 5, x_2= 14, x_3 = 5 and y1=5,y2=5,y3=17y_1 = 5, y_2 = 5, y_3 = 17.
a=15,b=12a = 15, b = 12 and c=9c = 9

(a,b)=(ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c)(a, b) = \left(\dfrac{ax_1 + bx_2 + cx_3}{a + b + c}, \dfrac{ay_1 + by_2 + cy_3}{a + b + c} \right)

=
(((15×5)+(12×14)+(9×5))15+12+9,((15×5)+(12×5)+(9×17))15+12+9)\left(\dfrac{((15 \times 5) + (12 \times 14) + (9 \times 5))}{15 + 12 + 9}, \dfrac{((15 \times 5) + (12 \times 5) + (9 \times 17))}{15 + 12 + 9} \right)

=
((75+168+45)36,(75+60+153)36)\left(\dfrac{(75 + 168 + 45)}{36}, \dfrac{(75 + 60 + 153)}{36} \right)

=
(28836,28836)\left(\dfrac{288}{36}, \dfrac{288}{36} \right)

(a,b)=(8,8)(a, b) = (8, 8)

(ii) (
c,dc, d) are the co-ordinates of the centroid of △PQR. (x1+x2+x33,y1+y2+y33)\left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3} \right)

(c,d)=(x1+x2+x33,y1+y2+y33)=(5+14+53,5+5+173)=(8,9)(c, d) = \left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3} \right) = \left(\dfrac{5 + 14 + 5}{3}, \dfrac{5 + 5 + 17}{3} \right) = (8, 9)

(iii) Area of the triangle with co-ordinates P
(5,5)(5, 5), Q (14,5)(14, 5), R (5,17)(5, 17)

⇒ Area of △PQR =
12×[(x1y2x2y1)+(x2y3x3y2)+(x3y1x1y3)]\dfrac{1}{2} \times [(x_1 y_2 - x_2 y_1) + (x_2 y_3 - x_3 y_2) + (x_3 y_1 - x_1 y_3)]

=
12×[(5×514×5)+(14×175×5)+(5×55×17)]\dfrac{1}{2} \times [(5 \times 5 - 14 \times 5) + (14 \times 17 - 5 \times 5) + (5 \times 5 - 5 \times 17)]

=
5454

Answer: (i)
(8,8)(8, 8), (ii) (8,9)(8, 9), (iii) 5454

4.3 Circle

Where (
a,ba, b) is the centre of a circle drawn with radius rr units, the equation of the circle is (xa)2+(yb)2=r2(x - a)^{2} + (y - b)^{2} = r^{2}

Example 18

A line passing through the centre of a circle intersects the circle at (2,11)(2, 11) and (4,3)(-4, 3). What is the equation of the circle?

Solution

The line segment connecting the two points
(2,11)(2, 11) and (4,3)(-4, 3) is the diameter. The mid-point of this line will be the centre of the circle.

Centre of the circle = Mid-point of the diameter =
(242,11+32)=(1,7)\left(\dfrac{2 - 4}{2}, \dfrac{11 + 3}{2} \right) = (-1, 7)

Length of the diameter =
(2(4))2+(11+3)2=36+64=10\sqrt{(2 - (-4))^{2} + (11 + 3)^2} = \sqrt{36 + 64} = 10

Length of the radius =
102=5\dfrac{10}{2} = 5

Equation of the circle =
(xa)2+(yb)2=r2(x+1)2+(y7)2=25(x - a)^{2} + (y - b)^{2} = r^{2} ⇒ (x + 1)^{2} + (y - 7)^{2} = 25

Answer:
(x+1)2+(y7)2=25(x + 1)^{2} + (y - 7)^{2} = 25

4.4 Area of Quadrilateral

Where the
44 vertices of a quadrilateral are (x1,y1),(x2,y2),(x3,y3),(x4,y4)(x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4),

Area of a Quadrilateral =
12(x1x3)(y2y4)(x2x4)(y1y3)\dfrac{1}{2}|(x_1 - x_3)(y_2 - y_4) - (x_2 - x_4)(y_1 - y_3)|

4.5 Area of Polygon

Where the n vertices of a polygon are
(x1,y1),(x2,y2),(x3,y3),...,(xn,yn)(x_1, y_1), (x_2, y_2), (x_3, y_3), ... , (x_n, y_n),

Area of a Polygon =
12(x1y2x2y1)+(x2y3x3y2)+...+(xny1x1yn)\dfrac{1}{2}|(x_1 y_2 - x_2 y_1) + (x_2 y_3 - x_3 y_2)+ ... +(x_n y_1 - x_1 y_n)|

Example 19

What is the area of the quadrilateral formed by the coordinates (2,3),(5,5),(2,4),(3,5)(-2, 3), (5, 5), (-2, -4), (3, -5)?

Solution


Upon plotting the points, we note that the order in which the points have to be arranged to form a quadrilateral is (2,3),(5,5),(3,5),(2,4)(-2, 3), (5, 5), (3, -5), (-2, -4).

Applying the formula,

Area =
12(x1x3)(y2y4)(x2x4)(y1y3)\dfrac{1}{2}|(x_1 - x_3)(y_2 - y_4) - (x_2 - x_4)(y_1 - y_3)|

12(2(3))(5(4))(5(2))(3(5))\dfrac{1}{2}|(-2 - (3))(5 - (-4)) - (5 - (-2))(3 - (-5))|

124556=1012=50.5\dfrac{1}{2}|- 45 - 56| = \dfrac{101}{2} = 50.5

Answer:
50.550.5





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