Area Of A Triangle In A Coordinate Plane
Let us consider a triangle whose base is parallel to x-axis.
Can you find its area?
Yes, it is a simple question to us as we know that area of triangle is given by the formula
Area
From the figure, it is clear that height of the triangle is 3 units and base is also 3 units.
Area of triangle ABC
square units
Now, consider the following figure
Now, can we calculate the area of ΔDEF? Here, we do not know the base and height. It is very difficult to find the base and height of ΔDEF but we can find the vertices of ΔDEF very easily.
The co-ordinates of D are (2, 5).
The co-ordinates of E are (5, 2).
The co-ordinates of F are (4, 7).
We can calculate the area of ΔDEF by a formula which involves the vertices of a triangle.
Let us derive that formula by considering any triangle, say ΔPQR, such that (x1, y1), (x2, y2), and (x3, y3) are the coordinates of the vertices P, Q and R respectively.
Here, line segments PA, QB and RC are the perpendiculars to X-axis from the vertices P, Q and R respectively. Therefore, PA || QB || RC and hence, quadrilaterals PQBA, PACR and QBCR are trapeziums.
From the figure, it can be seen that
Area of ΔPQR = Area of trapezium PQBA + Area of trapezium PACR – QBCR
We know that
Area of trapezium = (Sum of parallel sides × Perpendicular distance between parallel sides)
Therefore,
Area of ΔPQR = (QB + PA)BA + (PA + RC)AC – (QB + RC)BC
Now, let us find the area of ΔDEF using this formula.
Area of triangle DEF
Area of ΔDEF = 6 square units
In this way, we can calculate the area of a triangle in a coordinate plane by using this formula.
Can we have a triangle with area 0 square units? Let us see this.
Let us find the area of a triangle formed by the vertices (1, 4), (−1, 1), and (3, 7).
Now, area
The area of the triangle is 0. What does it mean?
It means that the three given points are collinear.
Thus, “if area of a triangle is zero, then its vertices will be collinear”.
Let us solve some more examples.
Example 1: Find the area of a triangle whose vertices are (−3, −2), (−2, 3), and (3, 4).
Solution:
Let A (−3, −2), B (−2, 3), and C (3, 4) be the vertices of the triangle.
By using the formula,Area of triangle ,
we obtain
Area of ΔABC
Neglecting the negative sign, we obtain
Area of triangle ABC = 12 square units
Example 2: Find the area of a quadrilateral whose vertices are (3, 7), (−5, 3), (−3, −2), and (5, −4) respectively? Solution:
Let ABCD be the quadrilateral as shown in the figure.
Draw a diagonal BD, which divides the quadrilateral into two triangles.
Now, area of quadrilateral ABCD = Area of ΔABD + Area of ΔBCD
Area of ΔABD
= 48 square units and area of ΔBCD
Area of quadrilateral ABCD = 48 + 18
= 66 square units
Thus, the area of quadrilateral ABCD is 66 square units.
Example 3: Find the value of x, if the points (0, 1), (x, −11), and (2, 9) are collinear.
Solution:
It is given that the points (0, 1), (x, −11), and (2, 9) are collinear. Thus, the area of triangle formed by these vertices will be zero.
Now, area of triangle
⇒ 0 =
⇒
⇒ 8x + 24 = 0
⇒ 8x = −24
⇒ x = −3
Thus, the value of x is −3.