A line is drawn through the point (1, 2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ , where O is the origin. If the area - Mathematics MCQ AIEEE 2012 2026 | ExamTest.live

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A line is drawn through the point (1, 2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is:

Solution & Step-by-step Explanation

Let the slope be m . The equation of the line through (1, 2) is y - 2 = m (x - 1) .To find P (x -intercept), set y = 0 : - 2 = m (x - 1) ⟹ x = 1 - \frac{2}{m} = \frac{m - 2}{m} .To find Q (y -intercept), set x = 0 : y - 2 = - m ⟹ y = 2 - m .Area A = \frac{1}{2} ∣ x \cdot y ∣ = \frac{1}{2} \frac{m - 2}{m} \cdot (2 - m) = \frac{1}{2} \frac{- ( m - 2 ) ^{2}}{m} = \frac{( m - 2 ) ^{2}}{- 2 m} (since m must be negative for a triangle in the first quadrant). A = \frac{m ^{2} - 4 m + 4}{- 2 m} = - \frac{m}{2} + 2 - \frac{2}{m} .To minimize A, set \frac{d A}{d m} = 0 : - \frac{1}{2} + \frac{2}{m ^{2}} = 0 ⟹ m^{2} = 4 ⟹ m = \pm 2 .For m < 0, m = - 2 .

Try it yourself before checking the explanation above.

A line is drawn through the point (1, 2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is:

-\frac{1}{4}

-4

-2

-\frac{1}{2}

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