Question:

Area of the triangle with vertices $(- 2, 2), (1, 5) $and $(6, - 1)$ is

Updated On: Jun 7, 2024
  • 15
  • 44625
  • 29/2
  • 33/2
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The Correct Option is D

Solution and Explanation

Area of triangle having vertices $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by
Area $=\frac{1}{2}\begin{vmatrix}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{vmatrix}$
$\therefore$ Required area $=\frac{1}{2}\begin{vmatrix}-2 & 2 & 1 \\ 1 & 5 & 1 \\ 6 & -1 & 1\end{vmatrix}$
$ =\frac{1}{2}[-2(5+1)-2(1-6)+1(-1-30)] $
$=\frac{1}{2}[-12+10-31] $
$=\frac{-33}{2} $
$\therefore$ Area $=\frac{33}{2} $ sq units
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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