Help!!!!!!!!!

Question

Find the equation of the straight line which is parallel to the line whose equation is 3x + 4y = 0, and which passes through the point of intersection of the lines x - 2y = a and x + 3y = 2a. Express the answer in the form Ax + By = C. Please show steps as well

Answer 1

Brad,

Parallel lines have the same slope. That means that the ratio of change in y to change in x is the same. With lines expressed in “standard form,” that means that Ax + By = C and Ax + By = D are parallel (because A and B determine the slope -A/B  and C and D determine the y-intercept as C/B  or D/B).

Now, where is the intersection of:
                            x - 2y = a
                           x + 3y = 2a
                     -----------------------              subtract equations
                                -5y = -a
                                   y = a/5

Substitute this y into the equation and solve for x:
            x - 2(a/5) = a
                        x = (7/5)a

The two equations are:
                      3x + 4y = 0
And                 3x + 4y = 3(7a/5) + 4(a/5) = 25a/5 = 5a
                       3x + 4y = 5a           (answer in standard form of equation)

Answer 2

First, we should probably find the point we need our line to go through - the intersection of the lines x - 2y= a and x+3y = 2a. So let's find where they intersect. We know that 2a= x+2y and a= x- 2y. Using that second equation, we know that 2a= 2x - 4y (just by multiplying by 2). But now we have two different equations for a! So since a must be a since number, let's set them equal.

 

x+3y = 2x - 4y

   3y  = x - 4y

   7y  = x

 

So x= 7y. Put this back into one of the equations, say a= x-2y. So we have

 

a = x - 2y

a = 7y - 2y

a = 5y

 

so y= a/5. Now we can use that to find x! 

 

a = x - 2(a/5)
a = x - (2a)/5
a+(2a)/5 = x

(5a)/5 + (2a)/5 = x

(7a)/5 = x

 

So we know the point of intersection is the point ( (7a)/5, a/5); that is, x= (7a)/5 and y=a/5.

 

Now we want the line to be parallel to 3x+4y=0. Parallel lines have the same slope. Let's solve for y to get the slope.

 

3x+4y=0

      4y= -3x

        y= -3/4 x

 

so the slope of this line is -3/4. So finally we know we need a line with a slope of -3/4 that goes through the point ((7a)/5, a/5). Lines look like y=mx+b, where m is the slope and b is the y-intercept. We know m=-3/4 because we know the slope. But what is b? Well, we know that if we plug in x=(7a)/5 and y=a/5 that this equation 'works', so let's use this to solve for b!

 

y=mx+b

y= -3/4 *x +b

a/5= -3/4*7a/5 +b

a/5= -21a/20 + b

a/5+21a/20 = b

4a/20 + 21a/20 =b

25a/20 =b

5a/4=b

 

So we know that m=-3/4 and b=(5a)/4. So the equation of the line we wanted is y=-3/4*x+(5a)/4.