###
**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 4

^{th}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 (x

_{1, }y_{1}), B (x_{2, }y_{2}) and C (x_{3, }y_{3}), then the**will be-***area of a triangle*
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

**if they lie on the same line. The three points, A (x***collinear*_{1, }y_{1}), B (x_{2, }y_{2}) and C (x_{3, }y_{3}), will be**if-***collinear*
The

**of a triangle is the point where all the three medians meet. The***centroid***of the triangle formed, by A (x***centroid*_{1, }y_{1}), B (x_{2, }y_{2}) and C (x_{3, }y_{3}), is –
The

**of a triangle is the point where all the perpendicular bisectors meet.The***circumcentre***‘O’ of the triangle formed, by A (x***circumcentre*_{1, }y_{1}), B (x_{2, }y_{2}) and C (x_{3, }y_{3}), is-
The

**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 (x***incentre*_{1, }y_{1}), B (x_{2, }y_{2}) and C (x_{3, }y_{3}), 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 3

^{rd}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 1

^{st}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 2

^{nd}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 3

^{rd}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

Remember to leave your answer in the comment section.

## 0 comments:

## Post a Comment