Question

Find a point on the line y = 4 that is equidistant from the two points (0,1) and (5,6)

Answer 1

Answer:

The point on the line $y=4$ that is equidistant from points (0,1) and (5,6) is (2, 4).

Step-by-step explanation:

Let be $y = 4$, $A = (0,1)$ and $B = (5,6)$. As we know that given function is a horizontal line, the condition of equidistance between that a point within that line and both points must be:

$r_{A/P} = r_{B/P}$

Where:

$r_{A/P}$ - Distance of point A with respect to P.

$r_{B/P}$ - Distance of point B with respect to P.

We expand this equivalence by Pythagorean Theorem:

$\sqrt{(x_{A}-x_{P})^{2}+(y_{A}-y_{P})^{2}} = \sqrt{(x_{B}-x_{P})^{2}+(y_{B}-y_{P})^{2}}$

$(x_{A}-x_{P})^{2} + (y_{A}-y_{P})^{2} = (x_{B}-x_{P})^{2}+(y_{B}-y_{P})^{2}$

$x_{A}^{2}-2\cdot x_{A}\cdot x_{P}+x_{P}^{2} + y_{A}^{2}-2\cdot y_{A}\cdot y_{P}+y_{P}^{2} = x_{B}^{2}-2\cdot x_{B}\cdot x_{P}+x_{P}^{2} + y_{B}^{2}-2\cdot y_{B}\cdot y_{P}+y_{P}^{2}$

$x_{A}^{2}-2\cdot x_{A}\cdot x_{P} + y_{A}^{2}-2\cdot y_{A}\cdot y_{P} = x_{B}^{2}-2\cdot x_{B}\cdot x_{P} + y_{B}^{2}-2\cdot y_{B}\cdot y_{P}$

And we get this expression:

$x_{A}^{2}+y_{A}^{2}-x_{B}^{2}-y_{B}^{2} - 2\cdot (x_{A}-x_{B})\cdot x_{P}-2\cdot (y_{A}-y_{B})\cdot y_{P} = 0$

If we know that $x_{A} = 0$, $y_{A} = 1$, $y_{P} = 4$, $x_{B} = 5$ and $y_{B} = 6$, the expression is reduced to this:

$0^{2}+1^{2}-5^{2}-6^{2}-2\cdot (0-5)\cdot x_{P} -2\cdot (1-6)\cdot (4) =0$

$10\cdot x_{P}-20=0$

The remaining component of the point within the line is:

$x_{P} = 2$

The point on the line $y=4$ that is equidistant from points (0,1) and (5,6) is (2, 4).