Problem 1 - A Complete Resource Book in Mathematics for JEE Main

Question

If one of the diagonals of a square is along the line \(x=\) \(y\) and one of its vertices is \((3,0)\), then its side through this vertex nearer to the origin is given by the equation. (A) \(y-3 x+9=0\) (B) \(3 y+x-3=0\) (C) \(x-3 y-3=0\) (D) \(3 x+y-9=0\)

Step-by-Step Solution

Answer

None of the given options provide the correct line equation, as these don't accommodate the provided conditions explicitly for side through vertex \((3,0)\).

Step 1: Understand the Given Conditions

We are given that one of the diagonals of the square is along the line \(x = y\) and one vertex is \((3, 0)\). A square has its diagonals perpendicular, meaning the other diagonal will have a line equation of \(x + y = c\).

Step 2: Determine the Center of the Square

Since the line \(x = y\) is a diagonal, the center of the square should also be a point on this diagonal. The center \((h, h)\) should also satisfy the equation of the other diagonal \(x + y = c\). Therefore, \(h + h = c\), so \(c = 2h\).

Step 3: Determine the Other Diagonal Equation

We need another equation for the other diagonal of the square. Let's use the given vertex \((3, 0)\). The center of the square is midpoint between any two opposite vertices. Using one vertex \((3,0)\), the midpoint formula gives the center \((h, h)\) as the average of the \(x\) and \(y\) coordinates of \((3, 0)\) and its opposite vertex \((x_2, y_2)\). Thus, \(\frac{3 + x_2}{2} = h\) and \(\frac{0 + y_2}{2} = h\).

Step 4: Establish Equations from Conditions

Equating \(h\) in terms of diagonal equations and the vertex \((3,0)\), and noticing the perpendicular nature of diagonals \(x = y\) and the other diagonal \(x + y = c\), we attempt finding equations for side lines. Substituting \(c = 3 + h\) and solving for lines close to origin gives the alternative line nearest to the origin passing through \((3, 0)\).

Step 5: Solve for the Line Equation

After matching the given options with equations derived from \((3,0)\) as a vertex on the line, the right equation close to line \(x = y\) is sought. Repeated element cross-verification suggests those \((3,0)\) passes through satisfies none regularly, effectively suggesting a scaled alteration with interchangeable constants, thus it slightly shifted. Verify option due to nearest default diagonal outcome.

Key Concepts

Coordinate Geometry

Coordinate Geometry is a part of mathematics that allows us to describe geometric figures algebraically by using a coordinate system. In a two-dimensional plane, we work primarily with the X (horizontal) and Y (vertical) axes.

By knowing the coordinates of key points, we can calculate distances, midpoints, and more, which simplifies understanding geometric shapes and their properties.

In the context of a square, such as in the original exercise, the concepts of symmetry and balance play a significant role:

• Symmetry ensures that the diagonals of the square intersect at right angles.
• The center of the square is equally distant from all four vertices.

By understanding these principles, we can better solve problems related to squares and other polygons in coordinate geometry.

Squares

Squares are special quadrilaterals that have equal sides and angles. In the context of coordinate geometry, we can use their properties to determine positions and equations fairly easily.

Key properties of squares include:

• All four sides are of equal length.
• All angles are right angles (90 degrees).
• Diagonals bisect each other perpendicularly and are equal in length.

In this exercise, knowing that one diagonal is along the line \(x = y\) enables us to understand that any line perpendicular to it must be in the form of \(x + y = \ ext{constant}\).
This knowledge helps us in deducing the equation of a side of the square.

Line Equations

In analytical geometry, line equations are fundamental in describing straight lines on a graph. They are usually expressed in the form \(Ax + By + C = 0\), where \(A\), \(B\), and \(C\) are constants that define the line's slope and position.

For a line that follows the path of \(x = y\), every point on this line has equal X and Y coordinates. Thus, any line perpendicular to it will have the form \(x + y = \ ext{constant}\).

In our problem, one of the square's diagonals is already known to be \(x = y\). With the vertex \((3, 0)\), finding a line near the origin involves manipulating the basic form of the equation to fit pass through this specific point and be consistent with the geometric properties of squares.
This exploration helps in accurately ascribing the correct option for the line equation in the exercise.