Question

If the distance between the two points given below is 2√29, then find the value of k, given that k > 0.  ​

Answer 1

Solution:-

➡️Distance between the above two points  =  2√29

↠√[(x2 - x1)² + (y2 - y1)²]  =  2√29

➡️Substitute (x1, y1)  =  (-7, 2) and (x2, y2)  =  (3, k).

↠√[(3 + 7)² + (k - 2)²]  =  2√29

↠√[10² + (k - 2)²]  =  2√29

↠√[100 + (k - 2)2]  =  2√29

➡️Square both sides. 

↠100 + (k - 2)²  =  (2√29)²

↠100 + k² - 2(k)(2) +  2²  =  2²(√29)²

↠100 + k² - 4k +  4  =  4(29)

↠k² - 4k + 104  =  116

➡️Subtract 116 from each side.

↠k² - 4k - 12  =  0

↠(k - 6)(k + 2)  =  0

↠k - 6  =  0  or  k + 2  =  0

↠k  =  6  or  k  =  -2

Because k > 0, we have

↪ k=6