Question

Find the value of m,if the distance between A(2,-1) and B(m,-5) be 4√2 units

$d =$
​

Answer 1

Answer:

$\tt m = 6$ or $\tt m = -2$

Step-by-step explanation:

The distance formula between two points $\tt (x_1, y_1)$ and $\tt (x_2, y_2)$ in a coordinate plane is given by:

$\tt d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

In this case, we have points A(2, -1) and B(m, -5), and the distance is given as $\tt 4\sqrt{2}$.

Substituting the coordinates into the distance formula, we get:

$\tt 4\sqrt{2} = \sqrt{(m - 2)^2 + (-5 - (-1))^2}$

Simplify the equation:

$\tt 4\sqrt{2} = \sqrt{(m - 2)^2 + (-4)^2}$

Square both sides to eliminate the square root:

$\tt (4\sqrt{2} )^2= (\sqrt{(m - 2)^2 + (-4)^2})^2$

$\tt 32 = (m - 2)^2 + 16$

Subtract 16 from both sides:

$\tt 32 -16= (m - 2)^2 + 16-16$

$\tt 16 = (m - 2)^2$

Now, take the square root of both sides, considering both the positive and negative roots:

$\tt 4 = m - 2 \quad \text{or} \quad -4 = m - 2$

For the positive root:

$\tt m - 2 = 4$

Add 2 to both sides:

$\tt m - 2 +2 = 4+2$

$\tt m = 6$

For the negative root:

$\tt m - 2 = -4$

Add 2 to both sides:

$\tt m - 2 +2= -4+2$

$\tt m = -2$

So, there are two possible values for $\tt m$:

$\tt m = 6$ or $\tt m = -2$.

Answer 2

Answer:

m = -2, 6

Explanation:

$\sf d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

• where $\sf (x_1, y_1)$ and $\sf (x_2, y_2)$ are points

Insert values:

$\sf 4\sqrt{2} =\sqrt{(2-m)^2+(-1-(-5))^2}$

simplify

$\sf 4\sqrt{2} =\sqrt{(2-m)^2+(4)^2}$

square both sides

$\sf (4\sqrt{2})^2 =(2-m)^2+(4)^2$

simplify

$\sf 32=(2-m)^2+16$

change sides

$\sf 32-16=(2-m)^2$

$\sf \pm \sqrt{16}=2-m$

$\sf \pm 4=2-m$

change sides

$\sf 4-2=-m, -4-2 = -m$

simplify

$\sf m = -2, m =6$

You can put the values inside the distance formula again to check if the answer comes 4√2 for checking and verifying the solution.