A line segment is bisected by a line with the equation # -7 y + 3 x = 2 #. If one end of the line segment is at #( 2 , 4 )#, where is the other end?

Answer 1

Infinite solutions: any of the points in the line #y=(3/7)x-26/7#, including the point #(110/29,-424/203)# when the segment is perpendicularly bisected.

First we should notice that the point (2,4) isn't in the line #3x-7y-2=0# since: #3*2-7*4-2=6-28-2=-24#

That being said, the problem admits of an infinite number of solutions.

Calling AB the line segment with M the midpoint we have: #A(2,4)# #M((2+x_B)/2,(4+y_B)/2)# #B(x_B,y_B)#
The orthogonal projection of A onto the line is also M when the segment is perpendicular bisected. The slope of the line #y=(3x-2)/7#is #k=3/7# So the slope of line perpendicular to the aforementioned is #p=-1/k=-7/3# And the equation of such a perpendicular line passing through A is #y-4=-7/3(x-2)# => #y=(-7x+14)/3+4# => #y=(-7x+26)/3#
Combining the equations of the 2 lines #(3x-2)/7=(-7x+26)/3# => #9x-6=-49+162# => #58x=168# => #x=84/29# #-> y=(252/29-2)/7# => #y=194/203# And #M(84/29,194/203)#
Finding B #x_M=1+x_B/2# => #84/29=1+x_B/2#=>#x_B=55/29*2=110/29# #y_M=2+y_B/2# =>#194/203=2+y_B/2# =>#y_B=-212/203*2=-424/203# So when the segment AB is perpendicularly bisected by the line mentioned in the problem #B(110/29,-424/203)#
But since it is not required that the bisection happens perpendicularly the set of possible points B lays in a line parallel to the bisecting line and passing through #(110/29,-424/203)#, itself one of the possible locations of B. Then the locus in which the point B can be located is #y+424/203=(3/7)(x-110/29)# #y=(3/7)x-330/203-424/203# #y=(3/7)x-26/7#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the other end of the line segment bisected by the line with the equation ( -7y + 3x = 2 ), given that one end is at the point ((2, 4)), we can follow these steps:

  1. Calculate the slope of the line given by ( -7y + 3x = 2 ).
  2. Since the line bisects the line segment, the slope of the line segment formed by the two endpoints is negative reciprocal to the slope of the given line.
  3. Use the point-slope form to find the equation of the line passing through the point ((2, 4)) with the calculated slope.
  4. Solve the system of equations formed by the given line ( -7y + 3x = 2 ) and the equation of the line passing through ((2, 4)).
  5. Find the intersection point, which represents the other end of the line segment.

Let's go through these steps:

  1. The given line equation is ( -7y + 3x = 2 ), rewrite it in slope-intercept form: [ y = \frac{3}{7}x - \frac{2}{7} ]
  2. The slope of the given line is ( \frac{3}{7} ), so the slope of the line segment formed by the endpoints is ( -\frac{7}{3} ).
  3. Using point-slope form with the point ((2, 4)) and slope ( -\frac{7}{3} ): [ y - 4 = -\frac{7}{3}(x - 2) ]
  4. Expand and simplify the equation: [ y - 4 = -\frac{7}{3}x + \frac{14}{3} ] [ y = -\frac{7}{3}x + \frac{26}{3} ]
  5. Now, solve the system of equations: [ \begin{cases} -7y + 3x = 2 \ y = -\frac{7}{3}x + \frac{26}{3} \end{cases} ]
  6. Substitute the expression for ( y ) from the second equation into the first equation and solve for ( x ): [ -7\left(-\frac{7}{3}x + \frac{26}{3}\right) + 3x = 2 ]
  7. Solve for ( x ): [ x = 6 ]
  8. Substitute the value of ( x ) back into the equation of the line to find ( y ): [ y = -\frac{7}{3}(6) + \frac{26}{3} = -14 + \frac{26}{3} = \frac{4}{3} ]

So, the other end of the line segment is at the point ( (6, \frac{4}{3}) ).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7