Area Of A Triangle In A Coordinate Plane

Let us consider a triangle whose base is parallel to x-axis.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_4f60d62d.png

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

Areahttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_m5bac3d62.gif

From the figure, it is clear that height of the triangle is 3 units and base is also 3 units.

Area of triangle ABChttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_m1038ab8d.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_m287c3fa2.gifsquare units

Now, consider the following figure

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_849d2cb.png

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.

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_08_16_48_08/4.png

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 = https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_08_16_48_08/mathmlequation4801835077493074452.png(Sum of parallel sides × Perpendicular distance between parallel sides)

Therefore,

Area of ΔPQR = https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_08_16_48_08/mathmlequation4801835077493074452.png(QB + PA)BA + https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_08_16_48_08/mathmlequation4801835077493074452.png(PA + RC)AC – https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_08_16_48_08/mathmlequation4801835077493074452.png(QB + RC)BC

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_08_16_48_08/mathmlequation730172673507020294.png

Now, let us find the area of ΔDEF using this formula.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_4704d6fb.png

Area of triangle DEF

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_m5db73526.gif

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, areahttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_12249059.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_m6be5af8b.gif

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.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_m3419e002.png

By using the formula,Area of triangle , https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_77cd9d0d.gif

we obtain

Area of ΔABC

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_9109b1f.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_m3afa44fe.gif

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.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_4d0dc8f9.png

Draw a diagonal BD, which divides the quadrilateral into two triangles.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_42dea5f1.png

Now, area of quadrilateral ABCD = Area of ΔABD + Area of ΔBCD

Area of ΔABDhttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_71c756e6.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_m6fb0e83b.gif

= 48 square units and area of ΔBCDhttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_494e0fc3.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_24745b1.gif

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 trianglehttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_m2d008add.gif

⇒ 0 =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_m2f2bb301.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/134/177/452/491/10.1.7.3.1_SMT_KSB_SS_html_m6387c096.gif

⇒ 8x + 24 = 0

⇒ 8x = −24

x = −3

Thus, the value of x is −3.