Question

SOLVE THE ABOVE QUESTION... ​

Answer 1

Step-by-step explanation:

firstly AB =

$ab = \sqrt{(2 - 1) {}^{2} + (4 - 7) {}^{2} }$

$\sqrt{1 + 9} = \sqrt{10}$

similarly BC =

$\sqrt{(k - 2) {}^{2} + (5 - 4) {}^{2} } \\ \sqrt{(k - 2) {}^{2} + 1}$

and also AC =

$\sqrt{(k - 1) {}^{2} + (5 - 7) {}^{2} } \\ \sqrt{(k - 1) {}^{2} + 4 }$

in first case <A = 90°

then BC is the hypotenuse

then by using pythagoras theorem

BC² = AB² + AC²

(k-2)² + 4 = 10 + (k-1)² + 4

k² +4 - 4k +4 - 4 - 10 = k² + 1 - 2k

k² - k² - 4k + 2k -10 + 4 = 0

- 2k - 6 = 0

2k = - 6

k = - 3

Similarly solve for other two