Question

Area of quadrilateral formed by the lines $x=-\frac{3}{2},y=-\frac{5}{2},x=-\frac{15}{4}$and $y=-\frac{3}{2}$ is equal to

• A. 50 sq.units
• B. 25 sq.units
• C. 21 sq.units
• D. 15 sq.units

Answer 1

The correct option is C 21 sq.units

Given: $x=\frac{-3}{2},y=\frac{-5}{2},x=\frac{15}{4}$ and $y=\frac{3}{2}$
The line in the form of x = a, is a line parallel to y-axis at a distance of a units in the right direction from the origin, if a > 0 and in the left direction from the origin, if a < 0.
Thus, the lines $x=\frac{-3}{2}$ and $x=\frac{15}{4}$ are parallel to y-axis at a distance of $\frac{3}{2}$ units on the left direction from the origin and at a distance of $\frac{15}{4}$ units on the right direction from the origin, respectively.
Similarly, an equation of the form y = b, is a line parallel to x-axis at a distance of b units on the upward direction from the origin, if b > 0 and in the downward direction from the origin, if b < 0.
Therefore, the lines $y=\frac{-5}{2}$ and $y=\frac{3}{2}$ are parallel to x-axis at a distance of $\frac{5}{2}$ units in the downward direction and at a distance $\frac{3}{2}$ of units in the upward direction.

Drawing the lines $x=\frac{-3}{2},y=\frac{-5}{2},x=\frac{15}{4}$ and $y=\frac{3}{2}$ in a coordinate plane, we get


Here, the point of intersections of lines as A, B, C and D.
ABCD is a rectangle such that distance between $x=\frac{-3}{2}$ and $x=\frac{-15}{4}$ is the length and the distance between $y=\frac{-5}{2}$ and $y=\frac{3}{2}$ is the breadth of the rectangle.
Hence, the correct answer is option (3).