Question

ANOTHER important question, if you give proper answer i will mark you brainliest

Answer 1

$\tt {\pink{SOLUTION:}}$

The triangle is right-angled triangle. We can clearly apply Pythagoras theorem.

Let's draw the figure (NOT NECESSARY)

Use the distance formula to find out the distance.

$d = {\blue{\sqrt {\left( {x_1 - x_2 } \right)^2 + \left( {y_1 - y_2 } \right)^2 }}$

Distance between Q(0,3) and R(1,-4):

$d = \sqrt {(1 - 0)^2 + (-4 - 3)^2}$

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

$d = \sqrt {1 + 49}$

$d=\sqrt{50}$

Distance between P(k,2) and R(1,-4):

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

$d=\sqrt{k^2+1-2k+36}$

$d=\sqrt{k^2-2k+37}$

Distance between P(k,2) and Q(0,3):

$d = \sqrt {(k - 0)^2 + (0 - 3)^2}$

$d=\sqrt{k^2+9}$

Here, using Pythagoras theorem:

$\to {\pink{PR^2=PQ^2+QR^2}}$

$\to (\sqrt{k^2-2k+37})^2=(\sqrt{k^2+9})^2+(\sqrt{50})^2$

$\to k^2-2k+37=k^2+9+50$

$\to -2k=59-37$

$\to -2k=22$

$\to k=\frac{22}{-2}$

$\to k=-11$

Thus, k= -11.

${\pink{HOPE \:\: THIS \:\: HELPS \:\: :D}}$