Question

Please help me solve 41 and 42

Answer 1

N 41) Part a) Find the equation of BC

A(-2,-1)  B(0,3)

Step 1

Find the slope AB

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

Find the slope BC

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

the answer Part a) is

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

the answer part b) is

the coordinates of point C is (6,0)

N 42) Part a) Find the coordinates of P

we know that

the point P is the midpoint of BC

Step 1

Find the midpoint of BC

x-coordinate of P=(-7+3)/2=-2

y coordinate of P=(4+0)/2=2

coordinates of P is (-2,2)

therefore

the answer part a) is

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

the answer Part b) is

the equation of the line AP is y=2x+6