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Area of Triangle in Coordinate Geometry
In Geometry, a triangle is a threesided polygon that has three edges and three vertices. The area of the triangle is the space covered by the triangle in a twodimensional plane. The formula for the area of a triangle is (1/2) × base × altitude. Let's find out the area of a triangle in coordinate geometry.
What Is the Area of a Triangle in Coordinate Geometry?
Coordinate geometry is defined as the study of geometry using coordinate points. The area of a triangle in coordinate geometry can be calculated if the three vertices of the triangle are given in the coordinate plane. The area of a triangle in coordinate geometry is defined as the area or space covered by it in the 2D coordinate plane. Let us understand the concept of the area of a triangle in coordinate geometry better using the example given below,
Consider these three points: A(−2,1), B(3,2), C(1,5). If you plot these three points in the plane, you will find that they are noncollinear, which means that they can be the vertices of a triangle, as shown below:
The area covered by the triangle ABC in the xy plane is the region marked in blue. Now, with the help of coordinate geometry, we can find the area of this triangle. Let us learn more about it in the following section.
How Do You Calculate the Area of A Triangle in Coordinate Geometry?
In coordinate geometry, if we need to find the area of a triangle, we use the coordinates of the three vertices. Consider ▵ABC as given in the figure below with vertices A(x_{1}, y_{1}), B(x\(_2\), y\(_2\)), and C(x\(_3\), y\(_3\)). In this figure, we have drawn perpendiculars AE, CF, and BD from the vertices of the triangle to the horizontal axis. Notice that three trapeziums are formed: BAED, ACFE, and BCFD.
We can express the area of a triangle in terms of the areas of these three trapeziums.
Area(ΔABC) = Area(Trap.BAED) + Area(Trap.ACFE)  Area(Trap.BCFD)
Now, the area of a trapezium in terms of the lengths of the parallel sides (the bases of the trapezium) and the distance between the parallel sides (the height of the trapezium):
Trapezium Area = (1/2) × Sum of bases × Height
Consider any one trapezium, say BAED. Its bases are BD and AE, and its height is DE. BD and AE can easily be seen to be the y coordinates of B and A, while DE is the difference between the x coordinates of A and B. Similarly, the bases and heights of the other two trapeziums can be easily calculated. Thus, we have:
Area(Trap.BAED) = (1/2) × (BD + AE) × DE
= (1/2) × (y\(_2\) + y\(_1\)) × (x\(_1\) − x\(_2\))
Area(Trap.ACFE) = (1/2) × (AE + CF) × EF
= (1/2) × (y\(_1\) + y\(_3\)) × (x\(_3\) − x\(_1\))
Area(Trap.BCFD) = (1/2) × (BD + CF) × DF
= (1/2) × (y\(_2\) + y\(_3\)) × (x\(_3\) − x\(_2\))
The expression for the area of the triangle in terms of the coordinates of its vertices can thus be given as,
Area(ΔABC) = Area(Trap.BAED) + Area(Trap.ACFE)  Area(Trap.BCFD)
= (1/2) × [(y\(_2\) + y\(_1\)) × (x\(_1\) − x\(_2\))] + (1/2) × [(y\(_1\) + y\(_3\)) × (x\(_3\) − x\(_1\))]  (1/2) × [(y\(_2\) + y\(_3\)) × (x\(_3\) − x\(_2\))]
However, we should try to simplify it so that it is easy to remember.
For that, we simplify the product of the two brackets in each terms:
= (1/2) (x\(_1\)y\(_2\) − x\(_2\)y\(_2\) + x\(_1\)y\(_1\) − x\(_2\)y\(_1\)) + (1/2) (x\(_3\) y\(_1\) − x\(_1\)y\(_1\) + x\(_3\)y\(_3\) − x\(_1\)y\(_3\)) − (1/2)(x\(_3\)y\(_2\) − x\(_2\)y\(_2\) + x\(_3\)y\(_3\) − x\(_2\)y\(_3\))
Take the common term 1/2 outside the bracket.
=(1/2) (x\(_1\)y\(_2\) − x\(_2\)y\(_2\) + x\(_1\)y\(_1\) − x\(_2\)y\(_1\) − x\(_3\) y\(_1\) − x\(_1\)y\(_1\) + x\(_3\)y\(_3\) − x\(_1\)y\(_3\) − x\(_3\)y\(_2\) + x\(_2\)y\(_2\)  x\(_3\)y\(_3\) + x\(_2\)y\(_3\))
Thus,
Area(ΔABC) = (1/2){x\(_1\)(y\(_2\) − y\(_3\)) + x\(_2\)(y\(_3\) − y\(_1\)) + x\(_3\)(y\(_1\) − y\(_2\))}
As the area is always positive.
(ΔABC) = (1/2) x\(_1\)(y\(_2\) − y\(_3\)) + x\(_2\)(y\(_3\) − y\(_1\)) + x\(_3\)(y\(_1\) − y\(_2\))
This is a symmetric expression, and there is an easy technique to remember it, which we will now discuss as Determinants Method.
Area of a Triangle Using Determinant Method
To calculate the area of a triangle using determinants, we use the formula as shown below,
Area = 1/2 \(\begin{bmatrix}{{x_1}}&{{y_1}}&{{1}}\\{{x_2}}&{{y_2}}&{{1}}\\{x_3}&{y_3}&1\end{bmatrix}\)
Let us solve the above expression to obtain the formula for the area of a triangle using coordinates. We will solve the determinant along the first column.
Now, the first term in the expression for the area is \({x_1}\left( {{y_2}  {y_3}} \right)\). To obtain this, we solve determinants for the first term in the first column. Ignore the terms in the first row and column other than the first term and proceed according to the following visual representation (the cross arrows represent multiplication). Solving determinant we get, x\(_1\)(y\(_2\)  y\(_3\)).
The second term in the expression for the area is x\(_2\)(y\(_3\) − y\(_1\)). To obtain this, we solve determinant for the second term in the first column. Ignore the terms in the second row and first column other than the first term in the second column. Solving determinant, we get x\(_2\)(y\(_1\)  y\(_3\)) = x\(_2\)(\({y_3}  {y_1}\)):
Next, the third term in the expression for the area is \({x_3}\left( {{y_1}  {y_2}} \right)\). To obtain this, we solve determinant for the third term in the first column. Ignore the terms in the first row and third column other than the first term in the third column:
Finally, we add these three terms to get the area (and divided by a factor of 2, because we had this factor in the original expression we determined):
Area = (1/2) x\(_1\)(y\(_2\) − y\(_3\)) + x\(_2\)(y\(_3\) − y\(_1\)) + x\(_3\)(y\(_1\) − y\(_2\))
Note that we have put a modulus sign (vertical bars) around our algebraic expression, and removed the negative sign because the area is always positive, which we obtained in the original expression. So even if we get a negative value through the algebraic expression, the modulus sign will ensure that it gets converted to a positive value.
We can write the above expression for area compactly as follows:
\(A = \frac{1}{2}\;\left {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&{{1}}\\{{x_2}}&{{y_2}}&{{1}}\\{x_3}&{y_3}&1\end{array}} \right\)
Important Notes:
 The area of a triangle cannot be negative. In case we get the answer in negative terms, we should consider the numerical value of the area, without the negative sign.
 To find the area of a triangle in coordinate geometry, we need to find the length of three sides of a triangle using the distance formula.

If three points A(x\(_1\),y\(_1\)), B(x\(_2\),y\(_2\)), and C(x\(_3\),y\(_3\)) are collinear, then x\(_1\)(y\(_2\) − y\(_3\)) + x\(_2\)(y\(_3\) − y\(_1\)) + x\(_3\)(y\(_1\) − y\(_2\)) = 0.
Challenging Questions
 If the distance between the points (2, 3) and (1, q) is 5, find the values of q.
 What is the formula for the area of quadrilateral in coordinate geometry?
Examples on Area of a Triangle in Coordinate Geometry

Example 1: Consider a triangle with the following vertices: A(−1,2), B(2,3), C(4,−3). Find the area of this triangle in coordinate geometry?
Solution: To illustrate, we will calculate each of the three terms in the formula for the area separately, and then put them together to obtain the final value.
\(\left {\begin{array}{*{20}{c}}{  1}&2&4\\2&3&{  3}\\1&1&1\end{array}} \right\)
We will use the determinant formula to find the area of the given triangle.
Area of ΔABC = 1/2 \(\left {\begin{array}{*{20}{c}}{  1}&2&4\\2&3&{  3}\\1&1&1\end{array}} \right\)
Area of ΔABC = 1/2 1(3  (3))  2(2  (3)) + 4(2  3)
Area(ΔABC) = (1/2) (−6)  (10) + (−4) = (1/2) × 20 = 10 sq.units
∴ The area of a triangle is 10 unit square.

Example 2: Find the area of a triangle with the vertices: A(3,4), B(4,7), and C(6,−3).
Solution:
We have:
(ΔABC) = (1/2) x\(_1\)(y\(_2\) − y\(_3\)) + x\(_2\)(y\(_3\) − y\(_1\))+ x\(_3\)(y\(_1\) − y\(_2\))
(ΔABC) = (1/2) 3(7 − (−3)) + 4((−3) − (−4)) + 6(4 − (7)) = 1230 + 4 − 18 = (1/2) × 16 = 8sq.units

Example 3: Find the area of the triangle in coordinate geometry, whose vertices are: A(1,−2), B(−3,4), C(2,3)
Solution:
Area of ΔABC = 1/2 \(\left {\begin{array}{*{20}{c}}1&{2}&1\\{3}&4&1\\2&3&1\end{array}} \right\)
Area of ΔABC = (1/2) x\(_1\)(y\(_2\) − y\(_3\)) + x\(_2\)(y\(_3\) − y\(_1\))+ x\(_3\)(y\(_1\) − y\(_2\))
(1/2)1 × (4 − 3)  3 × ((3) + 2)) + 2(−2 − 4) = (1/2) 1 15 − 12 =13 sq.units
∴The area of a triangle is 13 sq.units
Practice Questions on Area of a Triangle in Coordinate Geometry
FAQs on Area of a Triangle in Coordinate Geometry
How Do You Find the Length of Side of a Triangle Using Coordinates?
The distance formula is used to find the length of a triangle using coordinates. Distance formula can be used to find the length of any side given the coordinates of the triangle's vertices.
What Is the Formula of the Area of a Triangle in Coordinate Geometry?
The formula of area of triangle formula in coordinate geometry the area of triangle in coordinate geometry is: A = (1/2) x\(_1\)(y\(_2\) − y\(_3\)) + x\(_2\)(y\(_3\) − y\(_1\)) + x\(_3\)(y\(_1\) − y\(_2\)), where (x\(_1\),y\(_1\)),(x\(_2\),y\(_2\)), and (x\(_3\),y\(_3\)) are the coordinates of vertices of triangle.
How Do You Find the Area and Perimeter of a Triangle With Coordinates?
For the area and perimeter of a triangle with coordinates first, we have to find the distance between each pair of points by distance formula and then we apply the formula for area and perimeter.
How Do You Find the Area of a Triangle With 3 Coordinates?
Area of triangle with 3 points is: A = (1/2) x\(_1\)(y\(_2\) − y\(_3\)) + x\(_2\)(y\(_3\) − y\(_1\)) + x\(_3\)(y\(_1\) − y\(_2\)), where (x\(_1\),y\(_1\)),(x\(_2\),y\(_2\)), and (x\(_3\),y\(_3\)) are the coordinates of vertices of triangle.
How Do You Find the Area of Triangle Using Vertices?
The formula of the area of triangle in coordinate geometry is: A = (1/2)x\(_1\)(y\(_2\) − y\(_3\)) + x\(_2\)(y\(_3\) − y\(_1\)) + x\(_3\)(y\(_1\) − y\(_2\)), where (x\(_1\),y\(_1\)), (x\(_2\),y\(_2\)), and (x\(_3\),y\(_3\)) are the vertices of triangle.
How Do You Find the Area of an Isosceles Triangle Using Coordinates?
First, we use the distance formula to calculate the length of each side of the triangle. If two sides are equal then it's an isosceles triangle. We can apply the area of an isosceles triangle formula using the side lengths.
How Do You Find the Area of a RightAngled Triangle Using Coordinates?
First, we use the distance formula to calculate the length of each side of the triangle. If the squares of the smaller two distances equal the square of the largest distance, then these points are the vertices of a right triangle. or we can use the Pythagoras theorem. We can apply the area of a right triangle formula using side lengths.
How Do You Calculate the Area of a Triangle on a Graph?
The area of a triangle on a graph is calculated by the formula of area which is: A = (1/2) x\(_1\)(y\(_2\) − y\(_3\)) + x\(_2\)(y\(_3\) − y\(_1\)) + x\(_3\)(y\(_1\) − y\(_2\)), where (x\(_1\),y\(_1\)), (x\(_2\),y\(_2\)), and (x\(_3\),y\(_3\)) are the vertices of triangle.
How Do You Find the Missing Coordinate of a RightAngled Triangle?
We use the distance formula and Pythagoras theorem to calculate the missing coordinate of a rightangled triangle.
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