Question

O(0,0), B(-3,-1), C(-1,-3) are vertices of a triangle OBC. D is a point on OC and E is a point on OB. If the equation of DE is \(2x + 2y + \sqrt{2} = 0\), then the ratio in which the line DE divides the altitude of the triangle OBC is

• A. \(\sqrt{2} : 4\sqrt{2} + 2\)
• B. \(1 : 4\sqrt{2} + 1\)
• C. \(\sqrt{2} : 4\sqrt{2} - 2\)
• D. \(1 : 4\sqrt{2} - 1\)

Hint:

In coordinate geometry problems involving altitudes and ratios, finding the lengths of segments along the line from the vertex to the base is often the most direct method.

Answer 1

The Correct Option is D

Step 1: Find the equation of the Altitude from O to BC:
Slope of BC = \(\frac{-3 - (-1)}{-1 - (-3)} = \frac{-2}{2} = -1\). The altitude from O is perpendicular to BC. Slope of altitude = \(-\frac{1}{(-1)} = 1\). Equation of altitude (passing through origin): \(y = x\).
Step 2: Find the length of the Altitude (OM):
Let M be the foot of the perpendicular on BC. Equation of line BC: \(y - (-1) = -1(x - (-3)) \implies x + y + 4 = 0\). Intersection of altitude \(y=x\) and BC \(x+y+4=0\): \(x + x + 4 = 0 \implies 2x = -4 \implies x = -2, y = -2\). \(M = (-2, -2)\). Length of Altitude \(OM = \sqrt{(-2-0)^2 + (-2-0)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}\).
Step 3: Find intersection of Altitude and line DE:
Line DE: \(2x + 2y + \sqrt{2} = 0\). Substitute \(y=x\) (equation of altitude): \(2x + 2x + \sqrt{2} = 0 \implies 4x = -\sqrt{2} \implies x = -\frac{\sqrt{2}}{4}\). Let this intersection point be \(P\). \(P = (-\frac{\sqrt{2}}{4}, -\frac{\sqrt{2}}{4})\). Distance \(OP = \sqrt{2 \left(-\frac{\sqrt{2}}{4}\right)^2} = \sqrt{2 \cdot \frac{2}{16}} = \sqrt{\frac{4}{16}} = \frac{1}{2}\).
Step 4: Calculate the Ratio:
The point P lies on the altitude OM. We need the ratio in which P divides OM. Distance \(OP = 0.5\). Distance \(PM = OM - OP = 2\sqrt{2} - 0.5\). Ratio \(OP : PM = 0.5 : (2\sqrt{2} - 0.5)\). Multiply by 2 to clear decimals: Ratio = \(1 : (4\sqrt{2} - 1)\).