Question

the distance between A(1, 3 )B (a ,7 )is 5 find the value of a

Answer 1

GIVEN.

A=(1,3). and B=(a,7)

AB=5cm

so using distance formula



AB=√(a-1)²+(7-3)²


===>√a²+1-2a+16


===>√a²-2a+17

now it is given AB= 5cm

so

5=√a²-a+17

===>25=a²-2a+17


====>8=a²-2a

====>a²-2a-8



====>a²-2a-8=0


===>a²-4a+2a-8

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

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


mean

a=-2 and a=4

Answer 2

$\underline \bold{Solution:-}$

We have to find out the distance between two points A(1,3) and B(a,7).

Distance between A and B = 5

$\sqrt{ {(a - 1)}^{2} + {(7 - 3)}^{2} } = 5\\ \\On \: squaring \: both \: sides \\ \\ {[\sqrt{ {(a - 1)}^{2} + {(7 - 3)}^{2} }] }^{2} = {5}^{2} \\ \\ {(a - 1)}^{2} + {4}^{2} = 25 \\ \\ {(a - 1)}^{2} + 16 = 25 \\ \\ {(a - 1)}^{2} = 25 - 16 \\ \\ {(a - 1)}^{2} = 9 \\ \\ a - 1 = \sqrt{9}$
a - 1 = ±3

On taking positive sign

a - 1 = 3

a = 3 + 1

a = 4

On taking negative sign

a - 1 = - 3

a = - 3 + 1

a = -2

So, the value of a is 4 and -2.