16. Find the distance between P (–2, 5) and the line with equation x – 3y + 4 = 0. (1 point)
17(sqrt10)/ 10
0
-17(sqrt10)/ 10
13(sqrt10)/ 10 ~
17. Find the distance between the lines with equations 5x + 12y = 12 and y = –x + 3. (1 point)
24/13 ~
48/13
-48/13
48/17
18. Find an equation of the line that bisects the acute angles formed by the lines with equations 2x + y – 5 = 0 and 3x – 2y + 6 = 0. (1 point)
(2SQRT13-3SQRT5)x+(SQRT13+2SQRT5)y-5SQ…
(-2SQRT13-3SQRT5)x+(-SQRT13+2SQRT5)y+5… ~
(-2SQRT13+3SQRT5)x+(-SQRT13-2SQRT5)y+5…
(-2SQRT13-3SQRT5)x+(-SQRT13+2SQRT5)y+5…
19. Find an equation of the line that bisects the acute angles formed by the lines with equations 4x + y + 3 = 0 and x + y – 2 = 0. (1 point)
(4SQRT2+SQRT17)x + (SQRT2+SQRT17)y + 3SQRT2 - 2SQRT17 = 0
(SQRT2+SQRT17)x - (4SQRT2+SQRT17)y + 3SQRT2 + 2SQRT17 = 0
(4SQRT2-SQRT17)x + (SQRT2+SQRT17)y + 3SQRT2 + 2SQRT17 = 0 ~
(SQRT2-SQRT17)x - (4SQRT2+SQRT17)y - 3SQRT2 - 2SQRT17 = 0
20. If 90° < < 180°, express cos in terms of tan . (1 point)
-SQRT(1/ 1 + tan^2 theta) ~
SQRT(1/ 1 + tan^2 theta)
SQRT(1 + tan^2 theta)
-SQRT(1 + tan^2 theta)
the one with ~ are my answers
16. To find the distance between a point and a line, we can use the formula for the distance between a point (x1, y1) and a line Ax + By + C = 0:
Distance = |Ax1 + By1 + C| / √(A^2 + B^2)
In this case, the equation of the line is x - 3y + 4 = 0, so the values of A, B, and C are A = 1, B = -3, and C = 4. The coordinates of the point P are (-2, 5), so x1 = -2 and y1 = 5.
Plugging these values into the formula, we get:
Distance = |1*(-2) + (-3)*5 + 4| / √(1^2 + (-3)^2)
Distance = |-2 - 15 + 4| / √(1 + 9)
Distance = |-13| / √10
Distance = 13 / √10
So the distance between P (-2, 5) and the line x - 3y + 4 = 0 is 13 / √10.
17. To find the distance between two lines, we can use the formula for the distance between parallel lines:
Distance = |C1 - C2| / √(A^2 + B^2)
In this case, the equations of the lines are 5x + 12y = 12 and y = -x + 3. We can rewrite the second equation as -x - y = -3, so the values of A, B, C1, and C2 are A = 5, B = 12, C1 = 12, and C2 = -3.
Plugging these values into the formula, we get:
Distance = |12 - (-3)| / √(5^2 + 12^2)
Distance = |15| / √(25 + 144)
Distance = 15 / √169
Distance = 15 / 13
So the distance between the lines 5x + 12y = 12 and y = -x + 3 is 15 / 13.
18. To find the equation of the line that bisects the acute angles formed by two lines, we need to find the slope of the bisector line. The slope of the bisector line is the negative reciprocal of the product of the slopes of the two given lines.
The equations of the lines are 2x + y - 5 = 0 and 3x - 2y + 6 = 0. We can rewrite these equations in slope-intercept form (y = mx + b) to find the slopes. The slopes of the given lines are m1 = -2 and m2 = 3/2.
The slope of the bisector line is -1 / (m1 * m2), which gives us -1 / (-2 * (3/2)), or 1/3.
Since the bisector line passes through the intersection point of the two given lines, we can find the coordinates of the intersection point by solving the system of equations formed by the given lines:
2x + y - 5 = 0
3x - 2y + 6 = 0
Solving this system, we find x = 1 and y = 3. So the coordinates of the intersection point are (1, 3).
Now we have the slope (1/3) and a point (1, 3) on the bisector line. We can use the point-slope form of a line to find its equation:
y - y1 = m(x - x1)
Plugging in the values, we get:
y - 3 = (1/3)(x - 1)
3y - 9 = x - 1
x - 3y = -8
So the equation of the line that bisects the acute angles formed by 2x + y - 5 = 0 and 3x - 2y + 6 = 0 is x - 3y = -8.
19. Similar to the previous question, we need to find the slope of the bisector line, which is the negative reciprocal of the product of the slopes of the given lines.
The equations of the lines are 4x + y + 3 = 0 and x + y - 2 = 0. We can rewrite these equations in slope-intercept form to find the slopes. The slopes of the given lines are m1 = -4 and m2 = -1.
The slope of the bisector line is -1 / (m1 * m2), which gives us -1 / (-4 * -1), or -1/4.
To find the intersection point of the two given lines, we solve the system of equations:
4x + y + 3 = 0
x + y - 2 = 0
Solving this system, we find x = 1 and y = -1. So the coordinates of the intersection point are (1, -1).
Using the point-slope form, we can find the equation of the bisector line:
y - (-1) = (-1/4)(x - 1)
y + 1 = -(1/4)x + 1/4
y = -(1/4)x + 1/4 - 1
y = -(1/4)x - 3/4
So the equation of the line that bisects the acute angles formed by 4x + y + 3 = 0 and x + y - 2 = 0 is y = -(1/4)x - 3/4.
20. Since 90° < θ < 180°, we know that the cosine of an angle in this range is negative.
Since tanθ = sinθ / cosθ, we can rearrange the equation to get cosθ = sinθ / tanθ.
Using the Pythagorean identity for trigonometric functions (sin²θ + cos²θ = 1), we can substitute sin²θ as (1 - cos²θ) in the equation. This gives us:
cosθ = (1 - cos²θ) / tanθ
Simplifying further, we have:
cosθ = 1/tanθ - cos²θ / tanθ
To find cosθ in terms of tanθ, we can substitute the value of tanθ in the equation.
cosθ = 1 / tanθ - cos²θ / tanθ
cosθ = 1 / (tanθ) - cos²θ / (tanθ)
cosθ = (1 - cos²θ) / (tanθ)
cosθ = -√(1 / (1 + tan²θ))
Therefore, cos in terms of tan for 90° < θ < 180° is -√(1 / (1 + tan²θ)).