Question

Find the value of a when the distance between the points (3,a)and(4,1) is root10

Answer 1

Let the given points are A( 3 , a ) and B ( 4, 1 ).


Then,


X1 = 3 , X2 = 4 and Y1 = a , Y2 = 1

We have :


AB = √10 => AB² = (√10)²



=> AB² = 10


=> ( 4 - 3 )² + ( 1 - a )² = 10




=> (1)² + (1 - a )² = 10


=> ( 1 - a )² = 10 - 1



=> ( 1 - a )² = 9



=> ( 1 - a )² = (3)²



=> 1 - a = +- 3



=> 1 - a = 3 or 1 - a = -3


=> -a = 3 - 1 or -a = -3 - 1



=> -a = 2 or -a = -4




=> a = -2 or a = 4.

Answer 2

Answer:

a=-2 and 4

Step-by-step explanation:

The given points are:

(3,a) and (4,1)

Now, using the distance formula and the given condition, we have

$\sqrt{(4-3)^2+(1-a)^2}=\sqrt{10}$

Squaring on both the sides, we get

$(4-3)^2+(1-a)^2=10$

Solving the above equation, we get

⇒$1+1+a^2-2a=10$

⇒$2+a^2-2a=10$

⇒$a^2-2a-8=0$

⇒$a^2+2a-4a-8=0$

⇒$a(a+2)-4(a+2)=0$

⇒$(a-4)(a+2)=0$

Thus, the values of a are -2 and 4.