Question

find the value of 'a' when the distance between the points (3,a) and (4,1) is √10....
please tell me the answer in noteebook​

Answer 1

Step-by-step explanation:

√10=√(4-3)^2+(1-a)^2

by squaring both side

10=1+1+a^2-2a

10=a^2-2a+2

a^2-2a-8=0

a^2-4a+2a-8a=0

a(a-4)+2(a-4) =0

(a-4)(a+2)= 0

a=. -2 or a=4

Answer 2

Answer:

let (x1,y1) & (x2,y2) is (3,a) & (4,1)

use midpoint formula

(x1+x2)/2 + (y1+y2)/2= rt10

(3+4)/2 + (a+1)/2= rt 10

7/2+ (a+1)/2 = rt10

(8+a)/2=rt10

squaring on both sides,

(8+a)^2/4=10

(8+a)^2=10*4=40

a^2+16a+64-40=0

a^2+16a+24=0

now use formula a= (-b+/_b^2-4ac)/2a

Step-by-step explanation:

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