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Problem 14

# The points $$A(x, y), B(y, z)$$ and $$C(x, x)$$ represents the vertices of a right angled triangle, if : (a) $$x=y$$ (b) $$y=z$$ (c) $$\mathrm{z}=x$$ (d) $$x=y=z$$

Expert verified
Among the given options, only for 'd', when $$x=y=z$$, the given points form the vertices of a right angled triangle.
See the step by step solution

## Step 1: Calculation of All Possible Distances

Find the distance between all pairs of points. These distances represent the sides of the triangle. The formula to find the distance between two points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is $$\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$ . Using this, we find the distances to be : - $$AB = \sqrt{(y-x)^2 + (z-y)^2}$$- $$AC = \sqrt{(x-x)^2 + (x-y)^2} = |x-y|$$- $$BC = \sqrt{(y-x)^2 + (x-z)^2}$$

## Step 2: Applying Pythagorean Theorem

For the triangle to be right-angled, the square of the largest distance should be equal to the sum of the squares of the other two distances. Set up the equations corresponding to each of the options and see if any satisfy the Pythagorean theorem.

## Step 3: Check Value for Each Option

Substitute each of the given options into your equations from step 2. You should find that none of them, except for the option 'd', satisfy the Pythagorean theorem. For $$x=y=z$$, the distances become:- $$AB = \sqrt{(z-z)^2 + (z-z)^2} = 0$$- $$AC = |z-z| = 0$$- $$BC = \sqrt{(z-z)^2 + (z-z)^2} = 0$$For $$x=y=z$$, all vertices become the same point, meaning the triangle degenerates into a single point, which, technically, can be considered a right triangle with all sides equal to 0.

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