# coordinate geometry

• Apr 11th 2008, 08:07 AM
tyana21
coordinate geometry
the line y=2x+3 intersects the y-axis at A.The points BC on this line are such that AB=BC.The line through B perpendicular to AC passes through the point D(-1,6).Calculate

a)the eqn. of BD.
b)coordinates of B
c)coordinates of C

2)given that coordinates in each of the following,find the perpendicular bisector of AB.
a)A(1,2) B(3,4)
b)A(2,3) B(-1,9)

3)the line 3x+y=8 interscects the curve $3x^2$+ $y^2$=28 at A and B.Find,
a)the length of AB
b)the equation of the perpendicular bisector of AB.
• Apr 11th 2008, 08:25 AM
earboth
Quote:

Originally Posted by tyana21
the line y=2x+3 intersects the y-axis at A.The points BC on this line are such that AB=BC.The line through B perpendicular to AC passes through the point D(-1,6).Calculate

a)the eqn. of BD.
b)coordinates of B
c)coordinates of C

...

1. Draw a sketch of the given straight line, the point D and the perpendicular line to AC.

2. The slope of the perpendicular line must be $m=-\frac12$ . Use the point-slope-formula:

$y-6=-\frac12(x+1)~\implies~\boxed{y =-\frac12 x+\frac{11}2}$

3. The point B is the point of intersection between the given line and the perpendicular line:

$-\frac12 x+\frac{11}2=2x+3~\implies~x = 1~\implies~y=5$

Therefore the point B has the coordinates B(1, 5)

4. The point C has the coordinates C(2, 7) (You should now be able to explain how I got this result: Use the definition of slope and the equation of the given line)