Question

The following graph shows a portion of the circle x² + y2 = 9 in the second quadrant and a line segment with endpoints D (0, 3) and E (3,-1). חד 2 E -1 -3 (a) Find an equation for the line that would be the perpendicular bisector of the segment DE. (b) Find the coordinates of a point A, such that ADEA and DE - EA.

Answer 1

Question)

D = (0, 3) and E = (3, -1)

You need to find the equation of line that is perpendicular bisector of segment DE.

Perpendicular bisector of segment DE must pass through mid-point of segment DE.

Let B be the point which is mid-point of segment DE.

Using mid-point formula:

$\small B(x,y) = \left ( \frac{0+3}{2},\frac{3+(-1)}{2} \right )=\left ( \frac{3}{2},1 \right )$

Let y = mx + c be the equation of line that is perpendicular bisector of segment DE.

1 Slope of perpendicular bisector line(m) = slope of segent DE

Slope of segment DE = $\small (y_{2}-y_{1})/(x_{2}-x_{1})=(-1-3)/(3-0)=-4/3$

m = -4/3

Also y = mx + c passes through point (3/2 ,1)

$\small 1=-\frac{4}{3}\times \frac{3}{2}+c\Rightarrow 1 = -2+c\Rightarrow c=3$

Therefore, equation of line that is perpendicular bisector of segment DE is

$\small \therefore \boldsymbol{y = -\frac{4x}{3}+3}$

b)

Any point A on line y = -(4/3)x + 3 will form a triangle DEA such that DE = EA. But A cannot be coordinates of mid-point of line DE i.e B.

$\small A\neq \left ( \frac{3}{2},1 \right )$

$\small \textup{A can be any point except }\left ( \frac{3}{2},1 \right )$