Question

Hey


Find The Point on The X - axis which is equidistant from the A (-3,4) and B (1,-4) .



Plz Give Correct and explained answer.​

Answer 1

Step-by-step explanation:

topic :

• equidistant

given :

• the point on the x - axis

• equidistant from the = A (-3,4) and B (1,-4) .

to find :

• Find The Point on The X - axis

solution :

$\sqrt (a (-3))² + (0 − 4)² \\ \\ \\ \\ \\ \\ = \sqrt (a + 3)² + 16 \\ \\ \\ \\ \\ \\ \\ = (a − 1)² + (0-(-4))² \\ \\ \\ \\ = √(a − 1)² + 16 \\ \\ \\ \\ \\ \\ \\ = ⇒ (a + 3)² + 16 = (a − 1)² + 16 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ = a² + 6a+9 = a² - 2a + 1 \\ \\ \\ = ⇒8a = -8 \\ \\ \\ \\ \\ \\ \\ = ⇒a=-1 \\ \\ \\ \\ \\ \\ \\ = (a,0) = (-1,0)</p><p></p><p>$

thus , the answer is (a,0) = (-1,0)

Answer 2

Question:-

Find The Point on The X - axis which is equidistant from the A (-3,4) and B (1,-4) .

Given:-

● The point on X - axis.

● Equidistant from the A = (– 3,4) and B = (1,– 4).

Solution:-

$\sf \red{Distance \: Formula = \sqrt{{(x{ \sf{{_{2}}}}} \sf -x { \sf{{_{1}}}} \sf) {}^{2} + (y { \sf{{_{2}}}} -{y \sf{{_{1}}}} ) {}^{2} }}$

Let, the required point is C (x,0).

$\sf AC = \sqrt{x - ( - 3) {}^{2} + (0 - 4) {}^{2} }$

$\sf = \sqrt{(x + 3) {}^{2} + ( - 4) {}^{2} }$

$\sf = \sqrt{x {}^{2} + 9 + 6x + 16 }$

$\sf = \sqrt{x {}^{2} + 6x + 25 }$

$\sf BC = \sqrt{(x - 1) {}^{2} + (0 - ( - 4)) {}^{2} }$

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

$\sf = \sqrt{x {}^{2} - 2x + 17}$

According to the Question.

AC = BC

$\sf \sqrt{x {}^{2} + 6x + 25} = \sqrt{x {}^{2} - 2x + 17 }$

Squaring on both Sides.

$\sf {{ \cancel{x {}^{2} }}} + 6x + 25 = {{ \cancel{x {}^{2}}}} - 2x + 17$

$\sf6x + 2x = 17 - 25$

$\sf8x = - 8$

$\sf x = \: {{ \cancel{ \frac{ - 8}{8} }}} = - 1$

Answer:-

$\sf \fbox \red{Required Point is (–1,0)}$

Hope you have satisfied.⚘