Question

find the value of k for while the distance between points p(1,2) and q(-2,k) is 5 units.

Answer 1

Answer:

The value of k are 6 and -2 .

Step-by-step explanation:

Formula

$Distance\ Formula = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

The distance between points p(1,2) and q(-2,k) is 5 units.

Now putting the values in the formula

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

Taking square on the both sides of the above equation

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

$25=(-3)^{2}+(k-2)^{2}$

(a - b)² = a² + b² -2ab

$25=9+k^{2}+2^{2}-2\times k\times 2$

25 = 9 + k² + 4 - 4k

k² -4k -25 + 13 = 0

k² -4k -12 = 0

k² -6k + 2k - 12 = 0

k (k-6)+2(k-6) = 0

(k-6)(k+2)= 0

k = 6 , k= -2

Therefore the value of k are 6 and -2 .

Answer 2

Answer:

6 and -2

Step-by-step explanation:

the above is correct i will give it 10000