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**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.

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.

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.

move ahead you must revise the distance and section formulas, since they will

come handy in this blog.

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**Co-ordinate Geometry Formulas for a**

Triangle

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.

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

be-

points on the xy plane are- A (x1, y1), B (x2, y2)

and C (x3, y3), then the

**will***area of a triangle*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.

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

y1), B (x2, y2) and C (x3, y3),

will be

are said to be

**if they lie on the same line. The three points, A (x1,***collinear*y1), B (x2, y2) and C (x3, y3),

will be

**if-***collinear*
The

of a triangle is the point where all the three medians meet. The

of the triangle formed, by A (x1, y1), B (x2, y2)

and C (x3, y3), is –

*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

of a triangle is the point where all the perpendicular bisectors meet.The

‘O’ of the triangle formed, by A (x1, y1), B (x2, y2)

and C (x3, y3), is-

*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

of a triangle is the centre of the largest circle that be drawn in the

triangle. The

B (x2, y2) and C (x3, y3), is –

*incentre*of a triangle is the centre of the largest circle that be drawn in the

triangle. The

**‘I’ of the triangle formed, by A (x1, y1),***incentre*B (x2, y2) and C (x3, y3), is –

###
**Set 1: Co-ordinate**

Geometry Formulas of a Triangle

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

10

(ii)

12

12

(iii)

6

6

(iv)

8

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.

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-

4, y=3; converting the 3rd

equation in intercept form-

3x

+ 4y = 12

+ 4y = 12

3x/12

+ 4y/12 = 1

+ 4y/12 = 1

x/4

+ y/3 = 1

+ y/3 = 1

Plotting

all the three lines on the graph we get-

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.

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-

values we get-

Area of the Triangle

= ½ ( 4 ( 3 – 0) + 0 (

3 – 0) + 4 ( 0 – 3)) = 6 sq. units

= 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.

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-

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

sq. units

Therefore the area of the triangle formed

by the 3 lines is 6 sq. units.

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

160/13

(ii)

150/3

150/3

(iii)

140/3

140/3

(iv)

10

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-

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)

+ 7y =35 (i)

4x

+ 3y = 12 (ii)

+ 3y = 12 (ii)

y

= 0 (iii)

= 0 (iii)

Taking

equation (i) and (ii)-

equation (i) and (ii)-

(5x

+ 7y =35) x 4 => 20x + 28y = 140

+ 7y =35) x 4 => 20x + 28y = 140

(4x

+ 3y = 12) x 5 => 20x + 15y = 60

+ 3y = 12) x 5 => 20x + 15y = 60

13y

= 80

= 80

y

= 80/13

= 80/13

Substituting

this value in equation (i), we get-

this value in equation (i), we get-

5x

+ 7x 80/13 = 35

+ 7x 80/13 = 35

x

= -21/13

= -21/13

Now

we have the 1st vertex- (-21/13 , 80/13 )

we have the 1st vertex- (-21/13 , 80/13 )

Taking

equation (ii) and (iii)-

equation (ii) and (iii)-

4x

+ 3y = 12

+ 3y = 12

y

= 0

= 0

Substituting

the value of y in the (ii) equation we get –

the value of y in the (ii) equation we get –

4x

= 12

= 12

x

= 3

= 3

Now

we have the 2nd vertex- (3 , 0)

we have the 2nd vertex- (3 , 0)

Taking

equation (iii) and (i)-

equation (iii) and (i)-

y

= 0

= 0

5x

+ 7y =35

+ 7y =35

Substituting

the value of y in the (i) equation we get –

the value of y in the (i) equation we get –

5x

= 35

= 35

x

= 7

= 7

Now

we have the 3rd vertex- (7 , 0)

we have the 3rd vertex- (7 , 0)

Time

to substitute the values in the formula-

to substitute the values in the formula-

Substituting

values we get-

values we get-

Area

of the Triangle = ½ [ -21/13 (0) + 3 (0 – 80/13) + 7 (80/13)]

of the Triangle = ½ [ -21/13 (0) + 3 (0 – 80/13) + 7 (80/13)]

Area

of the Triangle = ½ [ 0 – 240/13 + 560/13]

of the Triangle = ½ [ 0 – 240/13 + 560/13]

Area

of the Triangle = ½ x 320/13

of the Triangle = ½ x 320/13

Area

of the Triangle = 160/13

of the Triangle = 160/13

Therefore

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

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

###
**Co-ordinate Geometry**

Formulas based Practice Questions

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 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)

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)

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)

Find the area of a square whose consecutive vertices are (11, 12) and (5, 4)

a) 13 b)

10 c) 100 d) 125

10 c) 100 d) 125

Remember to leave your answer in the comment section.

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