Question:

Draw a quadrilateral in the Cartesian plane, whose vertices are (-4, 5), (0, 7), (5, -5) and (-4, -2). Also, find its area.

Updated On: Oct 24, 2023
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Solution and Explanation

Let ABCD be the given quadrilateral with vertices A (-4, 5), B (0, 7), C (5,-5), and D (- 4, - 2).
Then, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA, the given quadrilateral can
be drawn as

quadrilateral in the Cartesian plane
To find the area of quadrilateral ABCD, we draw one diagonal, say AC.
Accordingly, area (ABCD) = area (ΔABC) + area (ΔACD)
We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is

\(\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\)

Therefore, area of ΔABC

\(=\frac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)|\, unit^2\)

\(=\frac{1}{2}|-4(12)+5(-2)|\,\,unit^2\)

\(=\frac{1}{2}|-48-10|\,\,unit^2\)

\(=\frac{1}{2}|-58|\,\,unit^2\)

\(=\frac{1}{2}×58\,\,unit^2\)

\(=29\,\,unit^2\)

Area of ΔACD

\(=\frac{1}{2}|-4(-5+)+5(-2-5)+5(-4)(5+5)|\, unit^2\)

\(=\frac{1}{2}|-4(3)+5(-7)-4|\,\,unit^2\)

\(=\frac{1}{2}|-12-35-40|\,\,unit^2\)

\(=\frac{1}{2}|-63|\,\,unit^2\)

\(=\frac{63}{2}\,\,unit^2\)

Thus, area (ABCD) \(=(29+\frac{29}{2})\,unit^2\,=\frac{58+63}{2}\,unit^2\,\frac{121}{2}\,unit^2\)

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Concepts Used:

Straight lines

A straight line is a line having the shortest distance between two points. 

A straight line can be represented as an equation in various forms,  as show in the image below:

 

The following are the many forms of the equation of the line that are presented in straight line-

1. Slope – Point Form

Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.

y – y0 = m (x – x0)

2. Two – Point Form

Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2)  are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes

The slope of P2P = The slope of P1P2 , i.e.

\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)

Hence, the equation becomes:

y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)

3. Slope-Intercept Form

Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by

y – c =m( x - 0 )

As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if

y = m x +c