Three Points have coordinates A(1,7), B(7,5) and C(0,-2). Find:
a) the equation of the perpendicular bisector of AB

b)the point intersection of this perpendicular bisector nd BC..

2. Originally Posted by Tawar
Three Points have coordinates A(1,7), B(7,5) and C(0,-2. Find:
a) the equation of the perpendicular bisector of AB

b)the point intersection of this perpendicular bisector nd BC..

This question doesn't belong here but in pre-algebra or (analytic) geometry...anyway:

a)(i) Calculate the middle point of the segment AB: it is M $(4,6)$.

(ii)Calculate the slope of the segment AB: $-\frac{1}{3}$

(iii) Calculate the equation of the line through M and perpendicular to AB: its slope must be $3$, and then the line is $y-6=3(x-4)$

b) Evaluate the line on which the segment AB is and then solve the system of linear equations determined by this equation and the one you found in (a-(iii))

Tonio

3. Hello, Tawar!

Three points have coordinates: . $A(1,7),\; B(7,5),\;C(0,-2)$

Find:

a) the equation of the perpendicular bisector of $AB$.

The perpendicular bisector of AB passes through the midpoint of AB: . $(4,6)$

Its slope is perpendicular to AB: . $m_{AB} \:=\:\frac{5-7}{7-1} \:=\:\frac{\text{-}2}{6} \:=\:-\frac{1}{3}$

The slope of the perpendicular bisector is: . $m \:=\:+3$

The equation is: . $y - 6 \:=\:3(x-4) \quad\Rightarrow\quad y \:=\:3x-6$ .[1]

b) the point intersection of this perpendicular bisector and BC.

Line BC has slope: . $m \:=\:\frac{\text{-}2-5}{0-7} \:=\:1$ . and passes through $C(0,-2)$

Its equation is: . $y - (-2) \:=\:1(x-0) \quad\Rightarrow\quad y \:=\:x-2$ .[2]

For the intersection, equate [1] and [2]: . $3x-6 \:=\:x-2 \quad\Rightarrow\quad x \:=\:2$

. . Substitute into [1]: . $y |:=\:3(2)-6 \quad\Rightarrow\quad y \:=\:0$

The intersection is: . $(2,0)$