7
Distance
12, p)
the point
between
locunits
of P Р
ounts (4, 5) and
the value​

1 thought on “7<br />Distance<br />12, p)<br />the point<br />between<br />locunits<br />of P Р<br />ounts (4, 5) and<br />the value​”

1. Answer:

   Solution

   The given points are P(-4, 7) and Q(2, -5).

   Then, x1=−4,y1=7andx2=2,y2=−5

   ∴PQ=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−−√

   ={2−(−4)}2+(−5−7)2−−−−−−−−−−−−−−−−−−−−−−√=62+(−12)2−−−−−−−−−−−√

   =36+144−−−−−−−√=180−−−√=36×5−−−−−√=65–√ units

   Reply