Question

the distance between the point (t,t) and 4,5 is t-3 unit then the value of t can be​

Answer 1

According to the question , the points (4,p) and (1,0) =5 i.e.,

(1−4)2+(0−p)2−−−−−−−−−−−−−−−√=5

[∵ distance between the points (x1,y1)and(x2,y2),d=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−−√]

⇒(3−)2+p2−−−−−−−−−−√=5

⇒9+p2−−−−−√=5

On squaring both the sides , we get

9+p2=25

⇒p2=16⇒p=±4

Hence, the required value of p is ±4.