If the extremities of the base of an isosceles triangle are the points $(2a, 0)$ and $(0, a)$ and the equation of one of the sides is $x = 2a$, then the area of the triangle, in square units, is :
- $\frac{5}{4}a^{2}$
- $\frac{5}{2}a^{2}$
- $\frac{25a^2}{4}$
- $5a^2$
The Correct Option is B
Solution and Explanation
$AB=\sqrt{4a^{2}+a^{2}}=\sqrt{5}a$
Now, $AC=BC \Rightarrow b=a^{2}\sqrt{4a^{2}+\left(b-a\right)^{2}}$
$b^{2}=4a^{2}+b^{2}+a^{2}-2ab$
$\Rightarrow 2ab=5a^{2} \Rightarrow b=\frac{5a}{2}$
$\therefore C=\left(2a, \frac{5a}{2}\right)$
Hence area of the triangle
$=\frac{1}{2}\begin{vmatrix}2a&0&1\\ 0&a&1\\ 2a&\frac{5a}{2}&1\end{vmatrix}=\frac{1}{2}\begin{vmatrix}2a&0&1\\ 0&a&1\\ 0&\frac{5a}{2}&0\end{vmatrix}$
$=\frac{1}{2}\times2a\left(-\frac{5a}{2}\right)=-\frac{5a^{2}}{2}$
Since area is always +ve, hence area
$=\frac{5a^{2}}{2}$ sunit
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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
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