The area (in square unit) of the triangle formed by $x+y+1=0$ and the pair of straight lines $x^{2}-3xy=2y^{2}=0$ is
- $\frac{7}{12}$
- $\frac{5}{12}$
- $\frac{1}{12}$
- $\frac{1}{6}$
The Correct Option is C
Solution and Explanation
$\Rightarrow \left(x-2y\right)\left(x-y\right)=0$
$\Rightarrow x=2y, x=y=0\,\,\,\ldots\left(i\right)$
Also, $x+y+1=0\,\,\,\,\ldots\left(ii\right)$
On solving Eqs. (i) and (ii), we get
$A\left(-\frac{2}{3}, -\frac{1}{3}\right), B\left(-\frac{1}{2},-\frac{1}{2}\right),C\left(0, 0\right)$
$\therefore$ Area of $\Delta ABC=\frac{1}{2}\begin{vmatrix}\frac{2}{3}&-\frac{1}{3}&1\\ \frac{1}{2}&-\frac{1}{2}&1\\ 0&0&1\end{vmatrix}$
$=\frac{1}{2}\left[\frac{1}{3}-\frac{1}{6}\right]=\frac{1}{2}\left[\frac{1}{6}\right]=\frac{1}{12}$
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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