Raul can hit a golf ball 26.4 yards. a.j. can hit a golf ball 10 times as far. how far can a.j. hit the ball

Because A.J. can hit it 10 times as far, multiply the distance Raul hits the ball by 10:

26.4 x 10 = 264

A.J. can hit the ball 264 yards.

Option B. -1.5x-3.5y=-31.5

Solution:

Equation of the lane passing through A and B: -7x+3y=-21.5

If we isolate y, and express the equation in the slope-intercept form y=mx+b, the coefficient of the variable "x" is the slope of this line:

Adding 7x both sides of the equation:

-7x+3y+7x=-21.5+7x

3y=7x-21.5

Dividing both sides of the equation by 3:

(3/3)y=(7/3)x-21.5/3

y=(7/3)x-21.5/3

The slope of this line is m1=7/3 (coefficient of variable x)

This line is perpendicular to the central line PQ, then the product of their slopes must be equal to -1.

Slope of AB: m1=7/3

Slope of PQ: m2

m1 m2 = -1

Replacing m1=7/3

(7/3) m2 = -1

Solving for m2: Multiplying both sides of the equation by 3/4:

(3/7)(7/3) m2 = (3/7)(-1)

m2 = - 3/7

According with the graph we know that the line PQ passes through the point P=(7,6)=(xp, yp)→xp=7, yp=6

Then, using the point-slope equation:

y-y1=m(x-x1); y1=yp=6, m=m2=-3/7, x1=xp=7

Replacing the values:

y-6=(-3/7)(x-7)

Multiplying both sides of the equation by 7:

7(y-6)=7(-3/7)(x-7)

7y-42=(-3)(x-7)

7y-42=-3x+21

Subtracting 7y and 21 both sides of the equation:

7y-42-7y-21=-3x+21-7y-21

-63=-3x-7y

-3x-7y=-63

Dividing all the terms by 2:

-(3/2)x-(7/2)y=-63/2

-1.5x-3.5y=-31.5