AREA OF TRIANGLE WITH 3 COORDINATES

Problem 1 :

Find the area of a triangle whose vertices are (3, 0), (7, 0) and (8, 4).

Solution:

The vertices of the given triangle A(3, 0), B(7, 0) and C(8, 4).

x1 = 3, y1 = 0, x2 = 7, y2 = 0, x3 = 8, y3 = 4

Area of a  triangle

=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]=12[3(0-4)+7(4-0)+8(0-7)]=12[-12+28-56]=12[-40]=-20

Area can't be negative.

Therefore, the area of the triangle is 20 square units.

Problem 2 :

The area of a triangle whose vertices are (5, 0), (8, 0) and (8, 4) (in sq.units) is 

A) 20    B) 12    C) 6

Solution:

The vertices of the given triangle A(5, 0), B(8, 0) and C(8, 4)

x1 = 5, y1 = 0, x2 = 8, y2 = 0, x3 = 8, y3 = 4

Area of a  triangle

=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]=12[5(0-4)+8(4-0)+8(0-0)]=12[-20+32+0]=12[12]=6 sq.units

So, option (C) is correct.

Problem 3 :

The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is (7/2, y), find the value of y.

Solution:

Let A(2, 1), B(3, -2) and C(7/2, y)

x1 = 2, y1 = 1, x2 = 3, y2 = -2, x3 = 7/2, y3 = y

Area of a triangle

Area ⧍ABC=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]5=122(-2-y)+3(y-1)+72(1+2)10=-4-2y+3y-3+21210=y+72y=10-72y=132

Problem 4 :

Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k) and (-k, -5) is 24 sq. units.

Solution:

The vertices of the given triangle A(1, -1), B(-4, 2k) and C(-k, -5)

x1 = 1, y1 = -1, x2 = -4, y2 = 2k, x3 = -k, y3 = -5

Area of a  triangle

Area ⧍ABC=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]=12[1(2k+5)+(-4)(-5+1)+(-k)(-1-2k)]=122k+5+16+k+2k2=122k2+3k+21Area of ⧍ABC=24 sq.units122k2+3k+21=242k2+3k+21=482k2+3k-27=02k2+9k-6k-27=0k(2k+9)-3(2k+9)=0(k-3)(2k+9)=0k=3 or k=-92

Problem 5 :

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5).

Solution:

The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).

Co ordinates of midpoint of AB=P(x1,y1)=2+42,1+32=62,42=(3,2)Co ordinates of midpoint of BC=Q(x2,y2)=4+22,3+52=62,82=(3,4)Co ordinates of midpoint of AC=R(x3,y3)=2+22,1+52=42,62=(2,3)Area ⧍PQR=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]=12[3(4-3)+3(3-2)+2(2-4)]=12[3(1)+3(1)+2(-2)]=12[3+3-4]=12(2)=1 sq.unit

Hence, the area of the required triangle is 1 sq. unit.

Problem 6 :

For what type of k, (k > 0), is the area of the triangle with vertices (-2, 5), (k, -4) and (2k + 1, 10) to 53 sq. units?

Solution:

The vertices of the given triangle A(-2, 5), B(k, -4) and C(2k + 1, 10)

x1 = -2, y1 = 5, x2 = k, y2 = -4, x3 = 2k + 1, y3 = 10

Area of a  triangle

Area ⧍ABC=12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]53=12[(-2)(-4-10)+k(10-5)+(2k+1)(5+4)]53=12[28+5k+9(2k+1)]28+5k+18k+9=10637+23k=10623k=69k=6923k=3

Recent Articles

  1. Solving Word Problems on Numbers Worksheet

    Feb 25, 24 07:44 AM

    Solving Word Problems on Numbers Worksheet

    Read More

  2. Solving Word Problems on Numbers

    Feb 24, 24 11:07 PM

    Solving Word Problems on Numbers

    Read More

  3. Reducing Ratios to Lowest Terms Worksheet

    Feb 24, 24 08:49 PM

    Reducing Ratios to Lowest Terms Worksheet

    Read More