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The lines with equations ax-2y=c and 2x+by=-c intersect at (1,-5)and are perpendicular. what is c?

User Steinybot
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To find the value of c, we can use the fact that the lines with equations ax - 2y = c and 2x + by = -c intersect at the point (1, -5) and are perpendicular to each other.

First, let's find the slopes of the two lines. The slope of a line in the form y = mx + b is equal to the coefficient of x (m). Therefore, the slope of the line ax - 2y = c is a/(-2), and the slope of the line 2x + by = -c is -2/b.

Since the lines are perpendicular, the product of their slopes should be -1. So we have:

(a/(-2)) * (-2/b) = -1

Simplifying this equation, we get:

ab = 4

Now, let's substitute the coordinates of the point (1, -5) into the equations of the lines to find the value of c.

For the equation ax - 2y = c, we have:

a(1) - 2(-5) = c

a + 10 = c

For the equation 2x + by = -c, we have:

2(1) + b(-5) = -c

2 - 5b = -c

Since the lines intersect at (1, -5), the values of c should be the same for both equations. Therefore, we can equate the expressions for c:

a + 10 = 2 - 5b

Now, we have two equations:

ab = 4 (from the perpendicular slope)

a + 10 = 2 - 5b (from equating c)

From the first equation, we can solve for a:

a = 4/b

Substituting this value into the second equation, we get:

4/b + 10 = 2 - 5b

Simplifying this equation, we have:

4 + 10b = 2b^2 - 10b

Moving all terms to one side, we get a quadratic equation:

2b^2 - 20b + 4 = 0

We can solve this equation for b using the quadratic formula:

b = (-(-20) ± sqrt((-20)^2 - 4(2)(4))) / (2(2))

Simplifying further, we get:

b = (20 ± sqrt(400 - 32)) / 4

b = (20 ± sqrt(368)) / 4

b = (20 ± 2sqrt(92)) / 4

b = 5 ± sqrt(92) / 2

Now, substituting the value of b back into the equation a = 4/b, we can find the value of a:

a = 4 / (5 ± sqrt(92) / 2)

a = 8 / (10 ± 2sqrt(92))

a = 4 / (5 ± sqrt(92))

Finally, substituting a and b into the equation ab = 4, we can find c:

(4 / (5 ± sqrt(92))) * (5 ± sqrt(92)) = 4

Simplifying this equation, we have:

4 = 4

Therefore, c can take any value since the equation is satisfied for any value of c.

User Carmensita
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