## How to find Area of a Triangle in coordinate geometry?

For a triangle with vertices at A (x1, y1), B (x2, y2) & C (x3, y3), Area is given by:

1/2 (x1 y2 – x2y1 + x2y3 – x3y2 + x3y1 – x1y3)

Trick to remember: Make two columns with 4 rows each. 1st row contains x¬1, x2, x3, x1 and 2nd row contains y1, y2, y3, y1. Draw criss-cross arrows from column left to right. Each arrow represents a multiplication. Arrows going down a row have to be added while arrows going up a row have to be subtracted.

## What else you can do inside qs leap ?

**This can also be written in form of a determinant:**

Area=

1 | 1 | 1 |

x1 | x2 | x3 |

y1 | y2 | y3 |

This formula is derived from the original area of triangle formula: ½ * base * height. Derivation is fairly simple but for our current discussion it is un-necessary. It is highly recommended to remember the formula in terms of coordinates in one form or another.

## What else you can do inside qs leap ?

## Centroid of a triangle:

### Q. What is a median and a centroid?

A. Median is a line that joins vertex of a triangle to mid-point of the opposite side. There are three medians in a triangle one from each vertex. All the medians pass through a common point called the centroid and it is conventionally represented as G as shown in the figure. Medians are represented by the red line.

Ref. to fig.

### Q. What are the coordinates of a centroid of a triangle?

A. For a triangle with vertices at A (x1, y1), B (x2, y2) & C (x3, y3); co-ordinates of the centroid, G (x, y) are given by the formula:

x = (x1 + x2 + x3) / 3

y = (y1 + y2 + y3) / 3

Fun Fact: Centroid is also the centre of mass of the uniform planar triangular object. Physics tells us that it is an important point. (I am so tempted to go on and write about centre of mass *sigh*)

## Circumcentre of a triangle:

### Q. What is circumcentre of a triangle?

A. Circumcentre is the point which is equidistant (at the same distance) from all the three vertices. It is the intersection point of the perpendicular bisectors of each side. It is the centre of circumcircle of the triangle. It is conventionally represented as C as shown in the figure.

### Q. What is a perpendicular bisector?

A. Perpendicular bisector is the line that is perpendicular to a given line segment and divides the it into two equal parts. The red lines in the figure are the perpendicular bisectors.

### Q. What is a circumcircle?

A. Circumcircle is the circle that passes through all the three vertices of the triangle

### Q. How to find the co-ordinates of a circumcentre?

A. For a triangle with vertices at P (x1, y1), Q (x2, y2) & R (x3, y3) here are the steps to find out the coordinates of the circumcentre.

**Step 1**: Find the mid-point PQ using section formula. (I ran out of variables so I am using (h, k). Those who know the significance of (h, k) please don’t be mad at me.)

Mid-point (h, k) = ((x1 + x2) / 2, (y1 + y2) / 2)

**Step 2**: Find the slope of the line perpendicular to PQ.

Slope will be m = – [(x2 – x1)/ (y2 – y1)]

**Step 3**: Find the equation of perpendicular bisector using point-slope form using mid-point and the slope m.

Equation will be:

(y – k) = m (x – h)

**Step 4**: Follow same steps to get equation of perpendicular bisector of QR.

**Step 5**: Solve the two equations simultaneously to get the coordinates of C.

## Orthocentre of a triangle:

### Q. What is an orthocentre of a triangle?

A. Orthocentre is a point where three heights of a triangle meet. As shown in the figure. Orthocentre is conventionally represented by H.

In the figure ABC is a triangle. AD, BE and CF are perpendiculars on BC, AC and AB respectively from vertices A, B, C respectively. These perpendiculars are the 3 heights of the triangle. The convergent point of these three lines is called the orthocentre.

### Q. What are the steps to find the orthocentre of a triangle?

A. There is a generic formula to find orthocentre of a triangle but it is not required for our purpose. The best way to find the orthocentre is to use the following steps. Ref. to figure

Step 1: Find slope of the line AB and BC:

mAB = (y1 – y2)/(x1 – x2 )

mBC = (y3 – y2)/(x3 – x2 )

Step 2: Find out slope of CF and AD. We know that CF is perpendicular to AB and AD is perpendicular to BC.

mCF = -1/mAB

and

mAD = -1/mBC

Step 3: Find equations of CF and AD from slope-point form of a line using the slopes above:

[slope point form for slope m and point (x1, y1) is given by : y – y1 = m(x – x1)]

Equation of CF (slope = mCF and point is C (x3, y3))

**y – y2 = mCF (x – x3) … 1**

Equation of AD (slope = mAD and point is A (x1, y1))

**y – y2 = mAD (x – x1) … 2**

Step 4: Solving equations 1 and 2 we get the co-ordinates of the orthocentre H.

**Next Topic Straight Lines**