Problem 90 - A Complete Resource Book in Mathematics for JEE Main

Question

A line is drawn through the point \((1,2)\) to meet the coordinate axes at points \(P\) and \(Q\) respectively such that it forms a triangle \(O P Q\), where \(O\) is the origin. If the area of the triangle \(O P Q\) is least, then the slope of the line \(P Q\) is (A) \(-\frac{1}{4}\) (B) \(-4\) (C) \(-2\) (D) \(-\frac{1}{2}\)

Step-by-Step Solution

Answer

The slope of the line is \\(-2\\) (Option C).

Step 1: Determine the Equation of the Line

The line passes through the point \((1,2)\) and has a slope \(m\). The general equation of the line is given by the point-slope form \(y - y_1 = m(x - x_1)\). For our case, this becomes \(y - 2 = m(x - 1)\), or equivalently, \(y = mx - m + 2\).

Step 2: Find Intercepts with Coordinate Axes

To find the intercept with the y-axis, set \(x = 0\) in the line equation: \(y = m(0) - m + 2 = -m + 2\). Thus, the y-intercept is \(Q(0, -m + 2)\). For the x-intercept, set \(y = 0\): \(0 = mx - m + 2\). Solving for \(x\), we get \(x = \frac{m - 2}{m}\). Hence, the x-intercept is \(P(\frac{m - 2}{m}, 0)\).

Step 3: Express Area of Triangle OPQ

The area of \(\triangle OPQ\) is given by the formula \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\). Here, the base \(OP\) is \(\left| \frac{m-2}{m} \right|\) and the height \(OQ\) is \(\left| -m+2 \right|\). Thus, the area is \(\frac{1}{2} \left( \frac{m-2}{|m|} \right) \left( |2 - m| \right)\).

Step 4: Simplify the Area Expression

Rewriting the area expression, we have \(\frac{1}{2} \cdot \frac{|2 - m| \cdot |m - 2|}{|m|}\). Since \(|m-2| = |2-m|\), the area simplifies to \(\frac{1}{2} \cdot \frac{(2-m)^2}{|m|}\).

Step 5: Determine Slope for Minimum Area

To minimize the area of \(\triangle OPQ\) given by \(A = \frac{(2-m)^2}{2|m|}\), we find the value of \(m\) that minimizes this expression. The minimum area occurs when the values inside the absolute deviations interact symmetrically around zero: \(|m| = |2-m|\). This implies \(m = -2\).

Step 6: Confirm Minimum Area Occurrence

Substitute \(m = -2\) into the simplified area expression: \(A = \frac{(2-(-2))^2}{2|-2|} = \frac{4^2}{4} = 4\), verifying that this produces a valid minimum result. Any deviation increases the calculated area.

Key Concepts

Area of Triangle

Understanding how to calculate the area of a triangle is essential, especially in coordinate geometry. When dealing with a triangle in a coordinate plane involving the origin and intercept points, the method is direct and applicable.

The formula for finding the area of a triangle located in a coordinate plane whose vertices are at the origin, and on the x and y axes, is quite interesting. Imagine a triangle formed by vertices corresponding to intercepts on the axes, and the origin at

• Base: Derived from the horizontal distance on the x-axis.
• Height: Derived from the vertical distance on the y-axis.

The conventional formula for the area of a triangle is: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]You use the coordinates of the intercepts to identify these measures. For this exercise, the base is the length along the x-axis where the line meets, and the height is the length along the y-axis that the line intercepts.

The calculation then becomes finding the product of these measures and adjusting your formula to provide 1/2 the product. The main takeaway is how to express these geometric properties from algebraic equations in coordinate terms.

Equation of Line

The equation of a line in coordinate geometry connects a given point's coordinates with a specific slope, dictating how the line rises or falls. To find the equation of a line through a particular point and with a known slope, we use the point-slope form of the line's equation.For instance, if a line passes through a known point \[ (x_1, y_1) \] with a slope \( m \), the equation is expressed as:\[ y - y_1 = m(x - x_1) \]Solving this gives the slope-intercept form, which relates directly to coordinate geometry by showing how the line progresses from its intersection with the y-axis. This line then extends through other points accordingly, mapping out its path in the coordinate system. By mathematically manipulating these values, you can precisely determine the intercepts, needed to solve or depict problems describing nexus points like triangle vertices in exercises.

Intercepts

Intercepts in coordinate geometry represent the points at which a line crosses the axes. These two points provide essential information regarding the line's position and can be calculated using the line equation properties.

• x-intercept: Occurs where the line crosses the x-axis.
• y-intercept: Where the line intersects the y-axis.

For instance, to find intercepts from a line equation, substitute values for one of the variables (either x or y) and solve for the other.- To find the y-intercept (point Q), set the x-coordinate as zero in the line equation: \[y = -m + 2 \]- To discover the x-intercept (point P), set the y-coordinate as zero and solve: \[0 = mx - m + 2 \]Recognizing and interpreting these intercepts is vital, especially in geometric problems where you use them to form shapes and solve for areas, distances, or other tasks. Intercepts provide the linkage to derive further properties like those discussed, such as triangle formations.