Question

Find the value or values of k for which the distance between the points a (k,-5),b(2,7) is 13 units

Answer 1

ab=13 
or,ab^2=(13)^2
or,(2-k)^2 + (7+5)^2=169
or,(2-k)^2 + (12)^2=169
or,(2-k)^2=169-144
or,(2-k)^2=25=(5)^2
or,2-k=+- 5 
or,2-k=5 or 2-k=-5
or,k=-3 or k=7

Answer 2

Answer:

In this question we will use distance formula:

i.e if we have to find distance b/w (a,b) and (c,d) then the formula used is :

\sqrt{(a - c) ^{2} + (b - d) ^{2} }

(a−c)

2

+(b−d)

2

now distance is already given in the question, so, we only need to put the values and get the answer.

13 = \sqrt{(k - 2)^{2} \: + (7 - ( - 5)) ^{2} }13=

(k−2)

2

+(7−(−5))

2

169 = 144 \: + {(k - 2)}^{2}169=144+(k−2)

2

|k - 2| = 5∣k−2∣=5

so there are two values of k , i.e there are two points in the plane having distance equivalent to 13 from (2,7).

k = 7 \: and \: - 3k=7and−3