Question

please answer this question ​

Answer 1

Answer:

The value of x-axis is (-5/2,0)

Explanation:

We have to find X axis so the coordinates will be (x, 0)

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3)

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5)

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3)

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0)

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0) AO = BO

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0) AO = BOAO = √(-2-x)² + (5-0)²

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0) AO = BOAO = √(-2-x)² + (5-0)² √4+x²+4x + 25

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0) AO = BOAO = √(-2-x)² + (5-0)² √4+x²+4x + 25 √ x²+4x+29... 1

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0) AO = BOAO = √(-2-x)² + (5-0)² √4+x²+4x + 25 √ x²+4x+29... 1BO = √ (2-x)² + (-3-0)²

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0) AO = BOAO = √(-2-x)² + (5-0)² √4+x²+4x + 25 √ x²+4x+29... 1BO = √ (2-x)² + (-3-0)² √4+x²-4x + 9

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0) AO = BOAO = √(-2-x)² + (5-0)² √4+x²+4x + 25 √ x²+4x+29... 1BO = √ (2-x)² + (-3-0)² √4+x²-4x + 9 √ x²-4x+13....2

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0) AO = BOAO = √(-2-x)² + (5-0)² √4+x²+4x + 25 √ x²+4x+29... 1BO = √ (2-x)² + (-3-0)² √4+x²-4x + 9 √ x²-4x+13....2 1 = 2

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0) AO = BOAO = √(-2-x)² + (5-0)² √4+x²+4x + 25 √ x²+4x+29... 1BO = √ (2-x)² + (-3-0)² √4+x²-4x + 9 √ x²-4x+13....2 1 = 2√x²+4x+29 = √x²-4x+9

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0) AO = BOAO = √(-2-x)² + (5-0)² √4+x²+4x + 25 √ x²+4x+29... 1BO = √ (2-x)² + (-3-0)² √4+x²-4x + 9 √ x²-4x+13....2 1 = 2√x²+4x+29 = √x²-4x+9 4x+29 = -4x+9

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0) AO = BOAO = √(-2-x)² + (5-0)² √4+x²+4x + 25 √ x²+4x+29... 1BO = √ (2-x)² + (-3-0)² √4+x²-4x + 9 √ x²-4x+13....2 1 = 2√x²+4x+29 = √x²-4x+9 4x+29 = -4x+9 4x+4x = 9-29

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0) AO = BOAO = √(-2-x)² + (5-0)² √4+x²+4x + 25 √ x²+4x+29... 1BO = √ (2-x)² + (-3-0)² √4+x²-4x + 9 √ x²-4x+13....2 1 = 2√x²+4x+29 = √x²-4x+9 4x+29 = -4x+9 4x+4x = 9-29 8x = -20

We have to find X axis so the coordinates will be (x, 0)(x, 0)is equidistant to (-2,5) and (2,-3) Let A = (-2,5) B = (2,-3) O = (x, 0) AO = BOAO = √(-2-x)² + (5-0)² √4+x²+4x + 25 √ x²+4x+29... 1BO = √ (2-x)² + (-3-0)² √4+x²-4x + 9 √ x²-4x+13....2 1 = 2√x²+4x+29 = √x²-4x+9 4x+29 = -4x+9 4x+4x = 9-29 8x = -20 x = -5/2