A line segment is bisected by a line with the equation # 5 y -4 x = 1 #. If one end of the line segment is at #(3 ,4 )#, where is the other end?

Answer 1

The other end is #(179/41,94/41)#

Let's rewrite the line's equation.

#5y-4x=1#
#5y=4x+1#
#y=4/5x+1/5#

The incline is

#m=4/5#
The slope of a line perpendicular is #m'=-5/4#
as #m*m'=-1#

The segment's equation is

#y-4=-5/4(x-3)#
#y=-5/4x+15/4+4=-5/4x+31/4#

The location where the lines converge

#y=4/5x+1/5=-5/4x+31/4#
#(4/5+5/4)x=-1/5+31/4#
#41/20x=161/20#
#x=151/41#
#y=4/5*151/41+1/5#
#y=(604+41)/205#
#y=645/205=129/41#
The point of intersection is #(151/41,129/41)#
Let the other end of the segment be #x_1,y_1#

Then,

#x_1-151/41=151/41-3#
#x_1=151/41+151/41-3=179/41#

and

#y_1-129/41=129/41-4#
#y_1=129/41+129/41-4=94/41#
The other end is #(179/41,94/41)# graph{(y-4/5x-1/5)(y+5/4x-31/4)((x-3)^2+(y-4)^2-0.01)((x-179/41)^2+(y-94/41)^2-0.01)=0 [-1.985, 9.116, -0.99, 4.56]}
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Answer 2

To find the other end of the line segment bisected by the line (5y - 4x = 1), we need to find the point of intersection between this line and the line segment. First, we'll determine the equation of the line segment using the given endpoint ((3, 4)) and the midpoint formula.

The midpoint formula states that the midpoint of a line segment with endpoints ((x_1, y_1)) and ((x_2, y_2)) is ((\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})).

Given one endpoint ((3, 4)) and assuming the other endpoint is ((x, y)), we find the midpoint.

Midpoint: ((\frac{3 + x}{2}, \frac{4 + y}{2}))

Now, since the line (5y - 4x = 1) bisects the line segment, the midpoint lies on it. Thus, we'll equate the coordinates of the midpoint to the equation of the line.

Equating the coordinates of the midpoint to the equation of the line, we have:

[\frac{3 + x}{2} = \frac{1}{4}(4x + 1)] [\frac{4 + y}{2} = \frac{1}{5}(5y - 1)]

Now, solve these equations to find (x) and (y). Once you have (x) and (y), you'll have the coordinates of the other end of the line segment.

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Answer 3

The other end of the line segment is at the point (7, 2).

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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.

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