Question

10)Find a point on x-axis ,which is equidistant from A(4,-3) & B(0,11)​

Answer 1

Answer:

Let (x₁y₁) be A(4,-3) and (x₂y₂) be the midpoint be P(x,0)

That is, x₁=4; y₁= -3; x₂=x; y₂=0.

Now, we Know that Distance of AP:

Use the distance formula:

=>√((x₂-x₁)² + (y₂-y₁)²)  

=>√{(x-4)²+(0-(-3))}

=>√(x²+16-8x+9)

=>√(x²+25-8x)

Now, let (x₁y₁) be B(0,11) and (x₂y₂) be the midpoint be P(x,0)

That is, x₁=0; y₁= 11; x₂=x; y₂=0.

Now, we Know that Distance of AB:

Use the distance formula:

=>√((x₂-x₁)² + (y₂-y₁)²)  

=>√{(x-0)²+(0-11)²}

=>√(x²+121)

But, it is given that AP=PB

∴√(x²+25-8x)=√(x²+121)

x²+25-8x=x²+121

-8x=121-25

-8x=96

x=96÷(-8)

x= -12

Thus, substitute the value of x in P(x,0)

⇒P(-12,0)

The point on x-axis ,which is equidistant from A(4,-3) & B(0,11)​ is -12,0.

HOPE THIS HELPS :D