Coordinate Geometry

Question

The line y= 2x+3 intersects the y-axis at A. The points B and C 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 equation of BD

(b)the coordinates of B

(c)the coordinates of C

Answer 1

y = 2x+3    This is slope intercept form so we have slope m = 2

and the y-intercept is at (0,3)

We know this intersects the y axis at A... That means that point A = (0,3)

 

If AB = BC then B is the midpoint of AC

 

(a)   We can find point B from the fact that a line perpendicular to AC going

       thru B passes thru point D at (-1,6)...

 

A perpendicular line to AC will have slope m  that is the negative reciprocal of the

slope of line AC .... negative reciprocal of 2 is -1/2

Thus we have a slope m = -1/2 and a point on the line (-1,6).. use point slope form

y-y₁=m(x-x₁)

y-6 = (-1/2)(x-(-1))

y-6 = -1/2(x+1)

y-6 = -1/2 x - 1/2

Multiply through by 2

2y-12 = -x -1

2y = -x + 11

 

The equation of BD is 2y = -x + 11

 

(b)  We now have the equation of a perpendicular line to AC passing through B

If we find the point where our original y = 2x+3 line touches the perpendicular

line 2y = -x+11 then we have point B

 

y = 2x+3         eqn 1

2y=-x+11        eqn 2

--------------

 

y = 2x + 3       eqn 1

4y=-2x+22      2(eqn 2)

-------------

5y = 25

y = 5

 

Substitute y = 5 back into y=2x+3

5 = 2x+3

2x = 2

x = 1

 

Point B is the point (1,5)

 

Since AB = BC then B is the midpoint of AC

 

A = (0,3)    B =  (1,5)

 

From A to B my x value went up 1, and my y value went up 2

From B if we go right 1 and up 2 we have point C at

C = (1+1, 5+2)

 

C = (2,7)