Question

If the locus of a point which is equidistant from the coordinate axes forms a triangle with the line \(y = 3\), then the area of the triangle is

• A. \(18 \)
• B. \(9 \)
• C. \(6 \)
• D. \(3 \)

Hint:

The locus of a point equidistant from the coordinate axes is the pair of lines \(y=x\) and \(y=-x\). To find the area of a triangle given its vertices, identify a suitable base and its corresponding perpendicular height. If one side is parallel to an axis, calculating base and height becomes straightforward.

Answer 1

The Correct Option is B

Step 1: Determine the locus of a point equidistant from the coordinate axes.
Let a point be \((x, y)\).
The distance from \((x, y)\) to the x-axis is \(|y|\).
The distance from \((x, y)\) to the y-axis is \(|x|\).
For the point to be equidistant from the coordinate axes, we must have \(|x| = |y|\).
This equality holds true for two lines:
1. \(y = x\) (points in the first and third quadrants where x and y coordinates are equal)
2. \(y = -x\) (points in the second and fourth quadrants where x and y coordinates are opposite)
These two lines pass through the origin \((0,0)\).
Step 2: Find the vertices of the triangle formed by these lines and the line \(y = 3\).
The triangle is formed by the lines \(L_1: y = x\), \(L_2: y = -x\), and \(L_3: y = 3\).
Vertex 1 (Intersection of \(L_1\) and \(L_3\)):
Substitute \(y = 3\) into \(y = x\), which gives \(x = 3\).
So, Vertex 1 is \((3, 3)\).
Vertex 2 (Intersection of \(L_2\) and \(L_3\)):
Substitute \(y = 3\) into \(y = -x\), which gives \(x = -3\).
So, Vertex 2 is \((-3, 3)\).
Vertex 3 (Intersection of \(L_1\) and \(L_2\)):
Set \(x = -x\), which implies \(2x = 0 \implies x = 0\). Since \(y=x\), \(y=0\).
So, Vertex 3 is \((0, 0)\) (the origin).
Step 3: Calculate the area of the triangle.
The vertices of the triangle are \(A(3, 3)\), \(B(-3, 3)\), and \(C(0, 0)\).
We can use the formula for the area of a triangle given its vertices.
Alternatively, observe that the base of the triangle can be taken along the line \(y = 3\).
The length of the base \(AB\) is the distance between \((3, 3)\) and \((-3, 3)\):
Base length \(b = |3 - (-3)| = |3 + 3| = 6\).
The height of the triangle is the perpendicular distance from the third vertex \(C(0, 0)\) to the line containing the base \(y = 3\).
Since \(y=3\) is a horizontal line, the height is the absolute difference in the y-coordinates:
Height \(h = |3 - 0| = 3\).
The area of the triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\):
\[ \text{Area} = \frac{1}{2} \times 6 \times 3 = 9 \] The final answer is \(\boxed{9}\).