A triangle $ABC$ has area $32$ sq units and its side $BC$ , of length $8$ units, lies on the line $\mathrm{x}=4$. Then the shortest possible distance between $A$ and the point $(0,0)$ is

We need to find where point A should be located to minimize its distance from the origin, given the constraints about the triangle.

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Since we know the area and one side length, we can find the height using the area formula:

Area of triangle = $\tfrac{1}{2} \times \text{base} \times \text{height}$

Given:

Area = 32 sq units

Base BC = 8 units

$32 = \tfrac{1}{2} \times 8 \times H$

$32 = 4H \rightarrow H = 8$ units

This means point A must be exactly 8 units away from line BC (measured perpendicular to BC).

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Since BC lies on the line $x = 4$, this is a vertical line.

For any point to be at a perpendicular distance of 8 from this vertical line, it must be 8 units away horizontally (left or right).

So point A can be located on either:

Line $x = 4 - 8 = -4$ (8 units to the left of BC)

Line $x = 4 + 8 = 12$ (8 units to the right of BC)

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Point A can be anywhere on one of these two vertical lines. Let's say A is at coordinates:

Either A = (-4, y) for some y-coordinate

Or A = (12, y) for some y-coordinate

Distance from A to origin (0,0):

If A = (-4, y): distance = $\sqrt{(-4)^2 + y^2} = \sqrt{16 + y^2}$

If A = (12, y): distance = $\sqrt{(12)^2 + y^2} = \sqrt{144 + y^2}$

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For both cases, the distance is minimized when $y = 0$.

Comparing minimum distances:

If A = (-4, 0): distance = $\sqrt{16} = 4$ units

If A = (12, 0): distance = $\sqrt{144} = 12$ units

Clearly, placing A at (-4, 0) gives the shortest distance.

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The shortest possible distance between A and the point (0,0) is 4 units.

Key Insight: When we have constraints on perpendicular distance, the optimal point lies on the line segment connecting the reference point (origin) to the constraint line, choosing the closer of the two possible positions.