A rectangle is drawn by lines x=0, x=2, y=0 and y=5. Points A and B lie on coordinate axes. If line AB divides the area of rectangle in 4:1, then the locus of mid-point of AB is?
Circle
Hyperbola
Ellipse
Straight line
The Correct Option is B
Approach Solution - 1
The correct answer is option (B): Hyperbola
A(2h,0), B(0,2k)
Area of \(\Delta\)OAB=8
\(\frac{1}{2}\times 2h\times 2k=8\)
\(hk=4\)
Locus of XY = 4
Approach Solution -2
Let the rectangle \( R \) be defined by the lines \( x = 0, \, x - 2y = 5 \). Thus, the vertices of the rectangle are at \( (0, 0), (5, 0), (0, 5), \) and \( (5, 5) \). The area of the rectangle is given by: \[ \text{Area of rectangle} = 5 \times 5 = 25 \]
Let the coordinates of \( A \) be \( A(\alpha, 0) \) and \( B(0, \beta) \), where \( \alpha \in [0, 5] \) and \( \beta \in [0, 5] \). We are told that the line segment \( AB \) divides the area of the rectangle in the ratio 4:1. This means that the area of triangle \( OAB \) is \( \frac{4}{5} \) of the total area of the rectangle. The area of triangle \( OAB \) is given by: \[ \text{Area of triangle } OAB = \frac{1}{2} \times \alpha \times \beta \] We are given that this area is \( \frac{4}{5} \) of the total area of the rectangle: \[ \frac{1}{2} \times \alpha \times \beta = \frac{4}{5} \times 25 = 20 \] Thus, we have the equation: \[ \alpha \times \beta = 40 \] Now, we know that the midpoint \( M \) of the line segment \( AB \) is given by: \[ M\left( \frac{\alpha}{2}, \frac{\beta}{2} \right) \] Substituting the equation \( \alpha \times \beta = 40 \), we get: \[ M\left( \frac{\alpha}{2}, \frac{\beta}{2} \right) \] This describes a locus of points that satisfies the equation of a hyperbola. Therefore, the midpoint \( M \) lies on a hyperbola.
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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