Find the equation of the line in general form with the given conditions:

Question

 a. m = ¼ , ( 4 , 2 )

b. x –intercept 5 , ( -2 , 3 )

c. y –intercept –2 , m = 5/3

 d. passing through A ( 4 , 5 ) and B ( -6 , -11 )

Answer 1

You should practice changing the equation of a line from General Form (Ax+By+C = 0) to slope-intercept form (y=mx+b) and back again.  This is because:

  -  in slope intercept form, m=slope and the point (0,b) is the y-intercept

  -  parallel lines all have the same slope (m)

  -  perpendicular lines have slopes that at the negative reciprocal of each other [that is, m and (-1/m)]

 

Once converted to slope-intercept form, the line Ax+By+C has slope of (-A/B).

 

a.    m = ¼ , ( 4 , 2 )

      m = -(1)/(-4)

  The equation of the line is:  x - 4y + C = 0    and it contains the point (4,2)

           4 - 4(2) + C = 0

            C = 4

      x - 4y + 4  = 0

 

b.    x –intercept 5 , ( -2 , 3 )

   Note:  x-intercept is the point where y=0

         Ax + B(0) + C = 0

              x = (-C/A)

            -C/A = 5

              C = -5A    [note: make C and A integers]

        Equation:     Ax + By -5A = 0

        Contains point (-2,3)

          A(-2) + B(3) -5A = 0

           7A = 3B          [So, set A=3 and B=7; C=-5A=-15]

        Equation:    3A + 7B -15 = 0

 

c.   y –intercept –2 , m = 5/3

       Using m= -A/B,   Equation is:  5x - 3y + C = 0

       Note: y-intercept is the place where x=0

             contains the point  (0,-2)

         5x -3y + C = 0

         5(0) -3(2) + C = 0

          C = 6

      Equation:   5x -3y + 6 = 0

      

 d. passing through A ( 4 , 5 ) and B ( -6 , -11 )

     CAUTION:  These are points A and B, not coefficients A and B,

          So, let's call them points P(4,5) and Q(-6,-11)

 

       Ax+By+C = 0 

          A(4) + B(5) + C = 0     [eq1]

         A(-6) + B(-11) + C = 0      [eq2]    

 

  Solve by elimination (but, you could use substitution):

        24A  +  30B   + 6C  =  0      [6*eq1]

       -24A   -  44B   + 4C  =  0      [4*eq2]

  -------------------------------------------------    [elimination; add equations]

                 - 14B    + 10C = 0

             10C = 14B

              5C = 7B              [so, let B=5, C=7]

  

Equation is:   Ax + 5y + 7 = 0     and use either point P or Q:

                  A(4) + 5(5) + 7 = 0

                   4A = -32

                     A = -8

 

Equation is:   -8x + 5y + 7 = 0

Equation is:    8x -5y -7 = 0      [multiply by (-1); by convention, coefficient of A should be positive]

 

It would be wise to check each of these answers.  Let's check (d) that points P(4,5) and Q(-6,-11) are on the line:

Check:

    Is   8(4) -5(5) - 7 = 0   ?

           32 - 25 - 7 = 0   ?

                    0 = 0   ?yes

    Is   8(-6) -5 (-11) -7 = 0   ?

              -48 + 55 - 7 = 0   ?

                    0 = 0   ?yes

Answer 2

From the point-slope form of a line,

 

y - y₁ = mx - mx₁

 

 

The general form is

 

-mx + y = -mx₁ + y₁

 

where:

m is the slope

(x₁ , y₁) is any of the given points the line passes through

 

 

Knowing this, can you find your lines for each part?

 

For part b, the x-intercept is the point (5, 0).

For part c, the y-intercept is the point (0, -2).