AREA OF TRIANGLE WITH COORDINATES

The area of a triangle with vertices A(x1, y1), B(x2, y2) and C(x3, y3)

Area of triangle = 12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

Problem 1 :

Find the area of the triangle ABC with A(1, -4) and the mid-points of sides through A being (2, -1) and (0, -1).

Solution :

The coordinates of the mid point of the line segment joining the points P(x1, y1) and Q(x2, y2) are

areaoftrianglewhenmidpointofsideisgiven
=(x1+x2)2'(y1+y2)2

Given, the midpoint A(1, -4) and (x, y) is (2, -1)

(2, -1)=(1+x)2'(-4+y)2

(1 + x)/2 = 2

1 + x = 4

x = 4 - 1

x = 3

(y - 4)/2 = -1

y - 4 = -2

y = -2 + 4

y = 2

The coordinates of B is (3, 2).

Given, the midpoint of A(1, -4) and C(a, b) is (0, -1).

(0, -1)=(1+a)2'(-4+b)2

(1 + a)/2 = 0

1 + a = 0

a = -1

(-4 + b)/2 = -1

b - 4 = -2

b = -2 + 4

b = 2

The coordinates of C are (-1, 2).

The area of a triangle with vertices A(x1, y1), B(x2, y2) and C(x3, y3)

=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

The area of a triangle with vertices A(1, -4), B(3, 2) and C(-1, 2)

=12[1(2-2)+3(2-(-4))+(-1)(-4-2)]=12[0+3(6)+(-1)(-6)]=12[18+6]=242=12 square units

Therefore, the area of the triangle ABC is 12 square units.

Problem 2 :

Find the value of m if the points (5, 1), (-2, -3) and (8, 2m) are collinear.

Solution:

Points A, B, C will be collinear if the area of ΔABC = 0

12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]=012[5(-3-2m)-2(2m-1)+8(1+3)]=0-15-10m-4m+2+32=0-14m-15+34=0-14m+19=0-14m=-19m=1914

Hence, the required value of m = 19/14.

Problem 3 :

Find the area of the triangle whose vertices are (-8, 4), (-6, 6) and (-3, 9)

Solution:

Let the vertices of triangle be,

A(x1, y1) = (-8, 4)  B(x2, y2) = (-6, 6) and C(x3, y3) = (-3, 9)

Area of triangle=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]Area of ⧍ABC=12[-8(6-9)-6(9-4)-3(4-6)]=12[24-30+6]=12(0)=0 square units

Hence, area of the triangle with vertices (-8, 4), (-6, 6) and (-3, 9) is 0 square units.

Problem 4 :

The points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a triangle ABC right angles at B. Find the values of a and hence the area of ΔABC.

Solution:

Given, the points A(2, 9), B(a, 5) and C(5, 5) are the vertices of right angle triangle whose angle B is right angle.

By distance formula,

AB=(2-a)2+(9-5)2=4+a2-4a+16=a2-4a+20BC=(a-5)2+(5-5)2=a2+25-10aCA=(5-2)2+(5-9)2=9+16=25=5

AB2 + BC2 = CA2

a2-4a+202+a2+25-10a2=52a2-4a+20+a2+25-10a=252a2-14a+20=02a2-7a+10=0a2-7a+10=0(a-2)(a-5)=0a=2 or a=5

Put first a = 2, In AB, BC and CA

Then,

AB=22-4×2+20=4-8+20=16=4BC=22-10×2+25=4-20+25=9=3

Now, put a = 5 and we get,

AB=52-4×5+20=25-20+20=25=5BC=52-10×5+25=25-50+25=0=0

Then a = 5 is not possible.

Hence a = 2 is possible for all value of right angle triangle.

Area of the right angles triangle ABC is

=AB×BC2=4×32=6 square units

Problem 4 :

A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ΔADE.

Solution:

Let fourth vertex of ABCD be D(x, y). We know that, the diagonals of a parallelogram bisects each other.

Mid point of BD = Mid point of AC

8+x2,2+y2=6+92,1+428+x2,2+y2=152,528+x2=152 and 2+y2=528+x=15 and 2+y=5x=7 and y=3

Thus, the required fourth vertex D is (7, 3).

Now, midpoint of DC 

=E7+92,3+42=E8,72

Area of ΔADE with vertices A(6, 1), D(7, 3) and E(8,7/2) is 

=1263-72+772-1+8(1-3)=12-3+352-16=12-32=|-34|=34 square units

Hence, the area of  ΔADE is 3/4 square units.

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