Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the value of a, if the distance between the points A (–3, –14) and B (a, –5) is 9 units.

Find the value of a, if the distance between the points A (–3, –14) and B (a, –5) is 9 units.

Answers (1)

best_answer

Solution.         Here, points are A(–3, –14) and B(a, –5)
Distance = 9
(x1, y1) = (-3, -14)                    (x2, y2) = (a, -5)
9= \sqrt{\left ( a+3 \right )^{2}+\left ( -5+14 \right )^{2}}
Squaring both sides we get
\left ( 9 \right )^{2}= \left ( a+3 \right )^{2}+\left ( 9 \right )^{2}
\left ( a+3 \right )^{2}= 0
\Rightarrow a+3= 0
a = -3
Value of a is -3

Posted by

infoexpert27

View full answer

Similar Questions

Latest Question