Question

find the value of K for which the distance between P3,2 Q 7,K is 5 find the value of K​

Answer 1

Answer:

The value of K is 6 and -2

Step-by-step explanation:

Step-by-step explanation:

Distance\ Formula

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}}5=

(−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}5

2

=(

(−2−1)

2

+(k−2)

2

)

2

25=(-3)^{2}+(k-2)^{2}25=(−3)

2

+(k−2)

2

(a - b)² = a² + b² -2ab

25=9+k^{2}+2^{2}-2\times k\times 225=9+k

2

+2

2

−2×k×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:

oflifififfidirdiogiridg

keodf