Question

Find the point on the x-axis which is equidistant From A(-3,4) and B(1,-4)​

Answer 1

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Answer 2

Step-by-step explanation:

the coordinates of point which is present on x axis is (x, 0)

the coordinates of point which is present on x axis is (x, 0)let P(x, 0) is equidistant point from A(-3, 4) and B(1, -4)

the coordinates of point which is present on x axis is (x, 0)let P(x, 0) is equidistant point from A(-3, 4) and B(1, -4)PA=PB

the coordinates of point which is present on x axis is (x, 0)let P(x, 0) is equidistant point from A(-3, 4) and B(1, -4)PA=PBsquaring on both sides

the coordinates of point which is present on x axis is (x, 0)let P(x, 0) is equidistant point from A(-3, 4) and B(1, -4)PA=PBsquaring on both sidesPA^2=PB^2

the coordinates of point which is present on x axis is (x, 0)let P(x, 0) is equidistant point from A(-3, 4) and B(1, -4)PA=PBsquaring on both sidesPA^2=PB^2[x-(-3) ]+(0-4) =(x-1) +[0-(-4) ]

the coordinates of point which is present on x axis is (x, 0)let P(x, 0) is equidistant point from A(-3, 4) and B(1, -4)PA=PBsquaring on both sidesPA^2=PB^2[x-(-3) ]+(0-4) =(x-1) +[0-(-4) ]x+3) +16=x^2-2x+1+16

the coordinates of point which is present on x axis is (x, 0)let P(x, 0) is equidistant point from A(-3, 4) and B(1, -4)PA=PBsquaring on both sidesPA^2=PB^2[x-(-3) ]+(0-4) =(x-1) +[0-(-4) ]x+3) +16=x^2-2x+1+16x^2+6x+9+16=x^2-2x+1+16

the coordinates of point which is present on x axis is (x, 0)let P(x, 0) is equidistant point from A(-3, 4) and B(1, -4)PA=PBsquaring on both sidesPA^2=PB^2[x-(-3) ]+(0-4) =(x-1) +[0-(-4) ]x+3) +16=x^2-2x+1+16x^2+6x+9+16=x^2-2x+1+166x+9=-2x+1

the coordinates of point which is present on x axis is (x, 0)let P(x, 0) is equidistant point from A(-3, 4) and B(1, -4)PA=PBsquaring on both sidesPA^2=PB^2[x-(-3) ]+(0-4) =(x-1) +[0-(-4) ]x+3) +16=x^2-2x+1+16x^2+6x+9+16=x^2-2x+1+166x+9=-2x+16x+2x=9-1

the coordinates of point which is present on x axis is (x, 0)let P(x, 0) is equidistant point from A(-3, 4) and B(1, -4)PA=PBsquaring on both sidesPA^2=PB^2[x-(-3) ]+(0-4) =(x-1) +[0-(-4) ]x+3) +16=x^2-2x+1+16x^2+6x+9+16=x^2-2x+1+166x+9=-2x+16x+2x=9-18x=8

the coordinates of point which is present on x axis is (x, 0)let P(x, 0) is equidistant point from A(-3, 4) and B(1, -4)PA=PBsquaring on both sidesPA^2=PB^2[x-(-3) ]+(0-4) =(x-1) +[0-(-4) ]x+3) +16=x^2-2x+1+16x^2+6x+9+16=x^2-2x+1+166x+9=-2x+16x+2x=9-18x=8x=8/8

the coordinates of point which is present on x axis is (x, 0)let P(x, 0) is equidistant point from A(-3, 4) and B(1, -4)PA=PBsquaring on both sidesPA^2=PB^2[x-(-3) ]+(0-4) =(x-1) +[0-(-4) ]x+3) +16=x^2-2x+1+16x^2+6x+9+16=x^2-2x+1+166x+9=-2x+16x+2x=9-18x=8x=8/8x=1

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