# A(1,7),B(7,5) and C(0,-2),find the point of intersection of BC with the perpendicular bisector of AB ?

Sep 21, 2015

$\left(2 , 0\right)$

#### Explanation:

This will be pretty long. I'm not sure if there's a shorter solution, but this is how I would solve it.

We will start by finding the equation of the perpendicular bisector of $\overline{A B}$.

Step 1 - Slope of the perpendicular bisector.

We will first find the slope of $\overline{A B}$ using the formula for slope.
$\frac{{y}_{1} - {y}_{2}}{{x}_{1} - {x}_{2}}$
$= \frac{\left(7\right) - \left(5\right)}{\left(1\right) - \left(7\right)}$
$= \frac{2}{- 6}$
$= - \frac{1}{3}$

We will now look for the slope of the perpendicular bisector. To do this, we will get the negative reciprocal of the slope of $\overline{A B}$
Slope of $\overline{A B}$: $\left(- \frac{1}{3}\right)$
Negative reciprocal: $3$

The slope of the perpendicular bisector is 3.

Step 2 - Point of intersection of $\overline{A B}$ and the perpendicular bisector.

Since this is a bisector we're talking about, it should go through the middle of $\overline{A B}$. To solve for the point of intersection, we will use the midpoint formula.
$\left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$
$\left(\frac{1 + 7}{2} , \frac{7 + 5}{2}\right)$
$\left(\frac{8}{2} , \frac{12}{2}\right)$
$\left(4 , 6\right)$

The point of intersection between $\overline{A B}$ and the perpendicular bisector is (4,6).

Step 3 - The equation of the perpendicular bisector.

Since we know the slope of the bisector and one point it intersects, we can write the equation of the line in point-slope form.
$y - {y}_{0} = m \left(x - {x}_{0}\right)$
$y - \left(6\right) = \left(3\right) \left[x - \left(4\right)\right]$
$y - 6 = 3 x - 12$
$y = 3 x - 12 + 6$
$y = 3 x - 6$

The equation of the perpendicular bisector is $y = 3 x - 6$.

We are done with the perpendicular bisector, so we can now proceed to finding the equation of the line that includes $\overline{B C}$.

Step 1 - Slope of $\overline{B C}$.
$\frac{{y}_{1} - {y}_{2}}{{x}_{1} - {x}_{2}}$
$= \frac{\left(5\right) - \left(- 2\right)}{\left(7\right) - \left(0\right)}$
$= \frac{7}{7}$
$= 1$

The slope of $\overline{B C}$ is 1.

Step 2 - Equation of the line including $\overline{B C}$.

We know that the line passes through (7,5) and (0,-2), so we can write the equation in point-slope form using either of the two points. I will use (0,-2) since it is simpler.
$y - {y}_{0} = m \left(x - {x}_{0}\right)$
$y - \left(- 2\right) = \left(1\right) \left[x - \left(0\right)\right]$
$y + 2 = x$
$y = x - 2$

The equation of the line including $\overline{B C}$ is $y = x - 2$.

FINAL STEP
Now that we have both the equations of the perpendicular bisector and the line including $\overline{B C}$, we can now look for the point of intersection using substitution.
$y = x - 2$
$3 x - 6 = x - 2$
$3 x - x = - 2 + 6$
$2 x = 4$
$x = 2$

To solve for the value of $y$, just substitute the value $2$ to either of the two equations.
$y = x - 2$
$y = \left(2\right) - 2$
$y = 0$

The point of intersection of $\overline{B C}$ and the perpendicular bisector is $\left(2 , 0\right)$.