TS EAMCET · Maths · Three Dimensional Geometry

If the points $A (1, 3, 5), B (2, 4, 6), C (4, 5, k)$ form a right angled triangle then the number of possible values of $k$ is

A $2$
B $3$
C $0$
D $1$

Verified Solution

Answer & Solution

Correct Answer

(A) $2$

Step-by-step Solution

Detailed explanation

Given, the points A1,3,5, B2,4,6,C4,5,k form a right angled triangle. Now, AB2=12+12+12=3 BC2=22+12+k-62=k2-12k+41 and CA2=32+22+k-52=k2-10k+38 If right angled at A, then AB2+AC2=BC2 ⇒3+k2-10k+38=k2-12k+41 ⇒k=0 If right angled at B, then AB2+BC2=AC2…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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If the points A1,3,5,B2,4,6,C4,5,k form a right angled triangle then the number of possible values of k is

Step-by-step Solution

If the points $A (1, 3, 5), B (2, 4, 6), C (4, 5, k)$ form a right angled triangle then the number of possible values of $k$ is