Question #acdcc

Line:
$y = 5 \cdot x - 13$

Circle:
${\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2} = 26$

If we want the intersection points, we substitute the line $y$ in the Circle equation:
${\left(x - 2\right)}^{2} + {\left(\left(5 \cdot x - 13\right) + 3\right)}^{2} = 26$
${\left(x - 2\right)}^{2} + {\left(5 \cdot x - 10\right)}^{2} = 26$
${\left(x - 2\right)}^{2} + {\left(5 \cdot \left(x - 2\right)\right)}^{2} = 26$
${\left(x - 2\right)}^{2} + 25 \cdot {\left(x - 2\right)}^{2} = 26$
$26 \cdot {\left(x - 2\right)}^{2} = 26$
${\left(x - 2\right)}^{2} = 1$
${x}^{2} - 4 \cdot x + 3 = 0$
${x}_{1} = 1$
${x}_{2} = 3$
${y}_{1} = 5 \cdot {x}_{1} - 13 = - 8 \to A = \left(1 , - 8\right)$
${y}_{2} = 5 \cdot {x}_{2} - 13 = 2 \to B = \left(3 , 2\right)$

Finding $M$:
If $M$ is the midpoint of $A B$,
$\vec{M} = \frac{\vec{A} + \vec{B}}{2}$
$\vec{M} = \left[1 + 3 , - 8 + 2\right] \cdot \frac{1}{2} = \left[2 , - 3\right]$
$M = \left(2 , - 3\right)$

Perpendicular line $y = m \cdot x + c$ :
As shown in
https://www.mathsisfun.com/algebra/line-parallel-perpendicular.html,
$m = - 0.2$
and $\left({P}_{0} = M\right)$
$\left(y - {y}_{0}\right) = m \cdot \left(x - {x}_{0}\right)$
$\left(y - \left(- 3\right)\right) = - 0.2 \cdot \left(x - 2\right)$
$y + 3 = - 0.2 \cdot x + 0.4$
$y = - 2.6 - 0.2 \cdot x$ is the equation of the perpendicular line to $A B$ that crosses it in $M$.

Hope it helps

May 21, 2015

Try this: 