Question

Find a if the distance of the point (4,1) from the point (3,a) is root 10

Answer 1

SOLUTION

$= > \sqrt{(4 - 3) {}^{2} + (1 - a) {}^{2} } = \sqrt{10} \\ = > 1 + (1 - a) {}^{2} = 10 \\ = > (1 - a) = \sqrt{9} \\ = > a = 1 - 3 = - 2 \\ or \\ a = 3 - 1 = 2$

Answer 2

ANSWER:-

Given:

The distance of the point (4,1) from the point (3,a) is √10.

To find:

Find the value of a.

Solution:

For given points;

⚫A (4,1)

⚫B (3,a)

x1 = 4, & x2= 3

y1 = 1, & y2= a.

We know that the formula of the distance, we get;

$d = \sqrt{(x2 - x1) {}^{2} + (y2 - y1) {}^{2} } \\ \\ = > \sqrt{(3 - 4) {}^{2} + (a - 1) {}^{2} } = \sqrt{10}$

On Squaring boths sides:

➨(3-4)² + (a-1)² = 10

➨(-1)² + (a-1)² = 10

➨1 + a² +1² -2a = 10

➨1 + a² + 1 -2a = 10

➨2 + a² - 2a = 10

➨2 + a² -2a -10=0

➨a² - 2a -8 =0

➨a² +2a -4a -8 =0

➨a(a+2) -4(a+2)=0

➨(a+2) (a-4) =0

➨a+ 2 = 0 or a-4=0

➨a= -2 or a= 4

Hence,

The value of a is -2 & 4.

Hope it helps ☺️