# Co-ordinate Geometry IV- Co-ordinate Geometry Formulas for Area of a Triangle and its Properties

### Now that we are in the final leg of our discussion in Co-ordinate Geometry, we will discuss the properties of a triangle and how to calculate the various parameters using co-ordinate geometry formulas.

Co-ordinate Geometry Formulas are a huge asset when working with the Cartesian system. This is our 4th blog in this series on co-ordinate geometry for SSCExams. In the previous three blogs we have discussed- the fundamentals of co-ordinate geometry, the basic co-ordinate geometry formulas and the various co-ordinate geometry formulas for equation of a straight line. In this blog we discuss co-ordinate geometry formulas for the area of a triangle and its different properties.
Before we move ahead you must revise the distance and section formulas, since they will come handy in this blog.

### Co-ordinate Geometry Formulas for a Triangle

In this section we will discuss co-ordinate geometry formulas to find the area of a triangle and its other properties when the co-ordinates for the 3 vertices are given.
If the three points on the xy plane are- A (x1, y1), B (x2, y2) and C (x3, y3), then the area of a triangle will be-
Sometimes, you may get a negative value as the area of a triangle. In such cases remember to take only the modulus or the absolute value.
Three points are said to be collinear if they lie on the same line. The three points, A (x1, y1), B (x2, y2) and C (x3, y3), will be collinear if-
The centroid of a triangle is the point where all the three medians meet. The centroid of the triangle formed, by A (x1, y1), B (x2, y2) and C (x3, y3), is –
The circumcentre of a triangle is the point where all the perpendicular bisectors meet.The circumcentre ‘O’ of the triangle formed, by A (x1, y1), B (x2, y2) and C (x3, y3), is-
The incentre of a triangle is the centre of the largest circle that be drawn in the triangle. The incentre ‘I’ of the triangle formed, by A (x1, y1), B (x2, y2) and C (x3, y3), is –

### Set 1: Co-ordinate Geometry Formulas of a Triangle

Example 1: Find the area of a triangle formed by points obtained by the equations => x = 4, y = 3 and 3x + 4y= 12.
(i) 10
(ii) 12
(iii) 6
(iv) 8

Solution 1:
These 3 equations represent a straight line, where x=4 is a straight line parallel to y-axis and y=3 is a straight line parallel to x-axis. Now the challenge in this question is that the vertices of the triangle have not been given directly but rather we have the equations of the lines. As per the co-ordinate geometry formulas of the area of a triangle, we need the vertices. So we need to solve the equations, get the vertices then replace them in the formula to get the answer.
x= 4,  y=3; converting the 3rd equation in intercept form-
3x + 4y = 12
3x/12 + 4y/12 = 1
x/4 + y/3 = 1

Plotting all the three lines on the graph we get-
Now we all the vertices for all the 3 points- (4,0), (0,3) and (4,3); we can easily find the area of the triangle using the appropriate formula from the list of co-ordinate geometry formulas.
Substituting values we get-
Area of the Triangle =  ½ ( 4 ( 3 – 0) + 0 ( 3 – 0) + 4 ( 0 – 3)) = 6 sq. units

Now this method can be very long and time taking. We can solve the same question by simply using the formula for the area of a triangle.
From the graph we can easily conclude that this is a right angles triangle, where the base is 4 and the height is 3. Substituting values in the above formula we get-
Area of the Triangle = ½ x 4 x 3 = 6 sq. units

Therefore the area of the triangle formed by the 3 lines is 6 sq. units.

Example 2: Find the area of a triangle formed by points obtained by the equations => 5x + 7y =35, 4x + 3y = 12 and the x-axis.
(i) 160/13
(ii) 150/3
(iii) 140/3
(iv) 10

Solution 2:
The three equations here represent three lines, so we have to find their points of intersection and substitute the values in the appropriate formula from the list of co-ordinate geometry formulas. Remember the equation of x-axis is y=0. So now we have 3 equations-
5x + 7y =35      (i)
4x + 3y = 12     (ii)
y = 0                  (iii)

Taking equation (i) and (ii)-
(5x + 7y =35) x 4 => 20x + 28y = 140
(4x + 3y = 12) x 5 => 20x + 15y = 60
13y = 80
y = 80/13
Substituting this value in equation (i), we get-
5x + 7x 80/13 = 35
x = -21/13
Now we have the 1st vertex- (-21/13 , 80/13 )

Taking equation (ii) and (iii)-
4x + 3y = 12
y = 0
Substituting the value of y in the (ii) equation we get –
4x = 12
x = 3
Now we have the 2nd vertex- (3 , 0)

Taking equation (iii) and (i)-
y = 0
5x + 7y =35
Substituting the value of y in the (i) equation we get –
5x = 35
x = 7
Now we have the 3rd vertex- (7 , 0)

Time to substitute the values in the formula-
Substituting values we get-
Area of the Triangle = ½ [ -21/13 (0) + 3 (0 – 80/13) + 7 (80/13)]
Area of the Triangle = ½ [ 0 – 240/13 + 560/13]
Area of the Triangle = ½ x 320/13
Area of the Triangle = 160/13

Therefore the area of the triangle is 160/13 sq. units

### Co-ordinate Geometry Formulas based Practice Questions

Question 1: A triangle is formed by the intersection of the lines 2x + 3y = 14, 4x – 5y = - 16 and the x – axis. Find the area of the triangle (in sq. units)
a) 20       b) 22       c) 25       d) 30 32.

Question 2: Find the area of the triangle formed by the points A (2, 4) B (4, 1) and C (-2, 1) (in sq. units)
a) 8       b) 12       c) 9       d) 10

Question 3: Find the area of the triangle formed by the points A (15, 15) B (16, 29) and C (50, 25) (in sq. units)
a) 280       b) 233       c) 245       d) 240

Question 4: Find the area of a square whose consecutive vertices are (11, 12) and (5, 4)
a) 13        b) 10      c) 100      d) 125