Find the slope of the line that passes through the given points (5,-1), (-7,2)

Len [333]

(6x - 5y + 4) dy + (y - 2x - 1) dx = 0

(6x - 5y + 4) dy = (2x - y + 1) dx

dy/dx = (2x - y + 1) / (6x - 5y + 4)

Let X = x - a and Y = y - b. We want to find constants a and b such that

dY/dX = (a rational function)

where the numerator and denominator on the right side are free of constant terms. Substituting x and y in the equation, we have

dY/dX = (2 (X + a) - (Y + b) + 1) / (6 (X + a) - 5 (Y + b) + 4)

dY/dX = (2X - Y + 2a - b + 1) / (6X - 5Y + 6a - 5b + 4)

Then we solve for a and b in the system,

2a - b + 1 = 0

6a - 5b + 4 = 0

==> a = -1/4 and b = 1/2

With these constants, the equation reduces to

dY/dX = (2X - Y) / (6X - 5Y)

Now substitute Y = VX and dY/dX = X dV/dX + V :

X dV/dX + V = (2X - VX) / (6X - 5VX)

The equation becomes separable after some simplification:

X dV/dX + V = (2 - V) / (6 - 5V)

X dV/dX = (2 - V) / (6 - 5V) - V

X dV/dX = (2 - V - (6 - 5V)) / (6 - 5V)

X dV/dX = (4V - 4) / (6 - 5V)

- (5V - 6) / (4V - 4) dV = 1/X dX

Integrate both sides:

-5/4 V + 1/4 ln|4V - 4| = ln|X| + C

Extract a constant from the logarithm on the left:

-5/4 V + 1/4 (ln(4) + ln|V - 1|) = ln|X| + C

-5/4 V + 1/4 ln|V - 1| = ln|X| + C

-5V + ln|V - 1| = 4 ln|X| + C

Get this back in terms of Y :

-5Y/X + ln|Y/X - 1| = 4 ln|X| + C

Now get the solution in terms of y and x :

-5 (y - 1/2)/(x + 1/4) + ln|(y - 1/2)/(x + 1/4) - 1| = 4 ln|x + 1/4| + C

With some manipulation of constants and logarithms, and a bit of algebra, we can rewrite this solution as

-5 (4y - 2)/(4x + 1) + ln|(4y - 4x - 3)/(4x + 1)| = 4 ln|x + 1/4| + 4 ln(4) + C

-5 (4y - 2)/(4x + 1) + ln|(4y - 4x - 3)/(4x + 1)| = 4 ln|4x + 1| + C

-5 (4y - 2)/(4x + 1) + ln|4y - 4x - 3| - ln|4x + 1| = 4 ln|4x + 1| + C

-5 (4y - 2)/(4x + 1) + ln|4y - 4x - 3| = 5 ln|4x + 1| + C

8 0

3 years ago

Could someone pls help me

babunello [35]

Answer: y = 90 x = 29

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the equation of the line that passes through (5, -1) and (-10, -10)

SashulF [63]

Answer:

y = 0.6x – 4

or

y = 3/5x -4

Step-by-step explanation:

3 0

3 years ago

In the expretion7to the third power-4*3+8 the firstt operation is

xxMikexx [17]

Answer:

7³

Step-by-step explanation:

Using PEMDAS, we see that E (which stands for exponents) comes before M (which stands for multiplication) and A (which stands for addition) so the first operation you should do is 7³.

8 0

3 years ago

Which points on the curve of x^2 - xy - y^2 = 5 have vertical tangent lines?

algol [13]

We need to differentiate this with respect to x to see if we can find an expression for the derivative of y at various points. That will be the slope of the tangent to the curve. Then we want to see where that derivative might be infinite -- i.e., where the tangent is vertical.

It's not written as a function, but it can still be differentiated using the chain rule:

x2 + xy + y2 = 3

(2x) + (x dy/dx + y dx/dx) + (2y dy/dx) = 0

(I used parentheses to show the differentiation of each term in the original equation.)

2x + x dy/dx + y + 2y dy/dx = 0

2x + y = -x dy/dx - 2y dy/dx

2x + y = dy/dx (-x -2y)

-(2x + y)/(x + 2y) = dy/dx

We have the derivative of y, but it's defined partly in terms of y itself. That's OK. Let's go on...

So where would the slope be infinite? That would happen when x + 2y = 0, or y = -x/2

Let's plug that in for y in the original equation to find points where that's the case.

x2 + xy + y2 = 3

x2 + x(-x/2) + (-x/2)2 = 3

x2 - x2/2 + x2/4 = 3

3x2 / 4 = 3

x2 = 4

x = ±2

So we have two x values where the tangent might be vertical. Let's plug them into the equation and see what the y values are. First x = 2...

x2 + xy + y2 = 3

4 + 2y + y2 = 3

y2 + 2y + 1 = 0

(y + 1)2 = 0

y = -1

So at the point (2, -1) the tangent is vertical.

Now try x = -2...

x2 + xy + y2 = 3

4 - 2y + y2 = 3

y2 - 2y + 1 =0

(y - 1)2 = 0

y = 1

So at the point (-2, 1) the tangent is vertical.

8 0

3 years ago