Question

The number of values of 'k' for which the points (-4,9,k), (-1,6,k), (0,7,10) form a right-angled isosceles triangle is

• A. 0
• B. 1
• C. 2
• D. 4

Hint:

When dealing with geometric conditions in 3D, first calculate the squared distances between the points. This avoids dealing with square roots. Then, apply the conditions (e.g., two sides equal for isosceles, Pythagoras' theorem for right-angled) to the squared lengths.

Answer 1

The Correct Option is C

Let the points be A(-4,9,k), B(-1,6,k), and C(0,7,10).
Let's calculate the squared lengths of the sides of the triangle ABC.
$AB^2 = (-1 - (-4))^2 + (6-9)^2 + (k-k)^2 = 3^2 + (-3)^2 + 0^2 = 9+9 = 18$.
$BC^2 = (0 - (-1))^2 + (7-6)^2 + (10-k)^2 = 1^2 + 1^2 + (10-k)^2 = 1+1+(10-k)^2 = 2 + (10-k)^2$.
$AC^2 = (0 - (-4))^2 + (7-9)^2 + (10-k)^2 = 4^2 + (-2)^2 + (10-k)^2 = 16+4+(10-k)^2 = 20 + (10-k)^2$.
For an isosceles triangle, two sides must be equal. We have three cases.
Case 1: $AB^2 = BC^2$. $18 = 2 + (10-k)^2 \implies (10-k)^2 = 16 \implies 10-k = \pm 4$. If $10-k=4$, $k=6$. If $10-k=-4$, $k=14$.
Case 2: $BC^2 = AC^2$. $2 + (10-k)^2 = 20 + (10-k)^2 \implies 2=20$, which is impossible. So this case is not possible.
Case 3: $AB^2 = AC^2$. $18 = 20 + (10-k)^2 \implies (10-k)^2 = -2$, which has no real solution for k.
So, for the triangle to be isosceles, $k$ must be 6 or 14. Now we check the right-angle condition for these values.
Check for $k=6$: $AB^2 = 18$. $BC^2 = 2 + (10-6)^2 = 2+4^2 = 18$. $AC^2 = 20 + (10-6)^2 = 20+4^2 = 36$. Check Pythagoras' theorem: $AB^2+BC^2 = 18+18 = 36 = AC^2$. The triangle is right-angled at B. This value of $k=6$ works.
Check for $k=14$: $AB^2 = 18$. $BC^2 = 2 + (10-14)^2 = 2+(-4)^2 = 18$. $AC^2 = 20 + (10-14)^2 = 20+(-4)^2 = 36$. The side lengths are the same as for $k=6$. So, $AB^2+BC^2=AC^2$ holds. The triangle is right-angled at B. This value of $k=14$ also works.
There are two possible values of 'k', which are 6 and 14.