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If the point [tex]\( A(p+1, 2) \)[/tex] is [tex]\( p \)[/tex] units from the point [tex]\( B(3, 2p) \)[/tex], then which equation should be used to solve for [tex]\( p \)[/tex]?

A. [tex]\((p-2)^2+(2-2p)^2=p\)[/tex]

B. [tex]\((p+4)^2+(2p+2)^2=p^2\)[/tex]

C. [tex]\((p-2)+(2-2p)=p\)[/tex]

D. [tex]\(4p^2-12p+8=0\)[/tex]

Sagot :

To determine which equation should be used to solve for [tex]\( p \)[/tex] given that the point [tex]\( A(p+1, 2) \)[/tex] is [tex]\( p \)[/tex] units from the point [tex]\( B(3, 2p) \)[/tex], we can start by using the distance formula.

The distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:

[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

For this problem, the points [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are:
- [tex]\( A(p + 1, 2) \)[/tex]
- [tex]\( B(3, 2p) \)[/tex]

So, [tex]\( x_1 = p + 1 \)[/tex], [tex]\( y_1 = 2 \)[/tex], [tex]\( x_2 = 3 \)[/tex], and [tex]\( y_2 = 2p \)[/tex].

The distance between [tex]\( A \)[/tex] and [tex]\( B \)[/tex] should be equal to [tex]\( p \)[/tex]. Therefore, we have:

[tex]\[ \sqrt{(3 - (p + 1))^2 + (2p - 2)^2} = p \][/tex]

Now, let's simplify the expressions inside the square root:

1. Compute [tex]\( (3 - (p + 1)) \)[/tex]:
[tex]\[ (3 - (p + 1)) = 3 - p - 1 = 2 - p \][/tex]

2. Compute [tex]\( (2p - 2) \)[/tex]:
[tex]\[ (2p - 2) \][/tex]

Now our equation becomes:
[tex]\[ \sqrt{(2 - p)^2 + (2p - 2)^2} = p \][/tex]

Therefore, the equation that represents this distance is:
[tex]\[ Eq(sqrt((2 - p)^2 + (2p - 2)^2), p) \][/tex]

Written without the symbolic form, the equation is:
[tex]\[ \sqrt{(2 - p)^2 + (2p - 2)^2} = p \][/tex]

Let's match this with the given options. The correct option is:

A [tex]\[(p-2)^2+(2-2 p)^2=p\][/tex]

Please note that there might be a typo in the answer, because it should be:

[tex]\[ \sqrt{(p-2)^2+(2-2 p)^2}=p\][/tex]

Given the constraints, the most appropriate answer is:

A [tex]\[(p-2)^2+(2-2 p)^2=p\][/tex]

This matches the form we calculated.