N 41) Part a) Find the equation of BC
A(-2,-1) B(0,3)
Step 1
<u>Find the slope AB</u>
m=(y2-y1)/(x2-x1)-----> m=(3+1)/(0+2)-----> mAB=2
we know that
If two lines are perpendicular. then the product of their slopes is equal to -1
mAB*mBC=-1
<u>Find the slope BC</u>
mBC=-1/2
Step 2
with the slope mBC and the point B find the equation of the line BC
y-y1=m*(x-x1)------> y-3=-0.5*(x-0)-----> y=-0.5x+3
therefore
<u>the answer Part a) is</u>
the equation of the line BC is y=-0.5x+3
Part b) Find the coordinates of C
Step 3
to find the coordinates of C find the x-intercept of the line BC
when y=0
y=-0.5x+3------> 0=-0.5x+3------> 0.5x=3-------> x=6
therefore
<u>the answer part b) is</u>
the coordinates of point C is (6,0)
<u>N 42) </u>Part a) Find the coordinates of P
we know that
the point P is the midpoint of BC
Step 1
<u>Find the midpoint of BC</u>
x-coordinate of P=(-7+3)/2=-2
y coordinate of P=(4+0)/2=2
coordinates of P is (-2,2)
therefore
<u>the answer part a) is</u>
The coordinates of P is (-2,2)
Part b) Find the equation of the line joining A and P
Find the slope AP
m=(2-8)/(-2-1)=2
with the slope and the point P find the equation of the line
y-y1=m*(x-x1)------> y-2=2*(x+2)------> y=2x+4+2------> y=2x+6
therefore
<u>the answer Part b) is</u>
the equation of the line AP is y=2x+6
