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
Were these right???????
16. To find the distance between point P(-2, 5) and the line with equation x - 3y + 4 = 0, we can use the formula for the distance between a point and a line. The formula is:
distance = |Ax + By + C| / sqrt(A^2 + B^2)
In this case, A = 1, B = -3, and C = 4. Plugging these values into the formula, we get:
distance = |1*(-2) + (-3)*5 + 4| / sqrt(1^2 + (-3)^2)
= |-2 - 15 + 4| / sqrt(1 + 9)
= |-13 + 4| / sqrt(10)
= |(-9)| / sqrt(10)
= 9 / sqrt(10)
So the distance between P (-2, 5) and the line x - 3y + 4 = 0 is 9 / sqrt(10), which is approximately 2.834.
17. To find the distance between the lines with equations 5x + 12y = 12 and y = -x + 3, we need to find the perpendicular distance between the two lines. We can find the perpendicular distance using the formula:
distance = |C1 - C2| / sqrt(A^2 + B^2)
Where the equations of the lines are in the form Ax + By + C = 0. In this case, the equations are already in this form.
For the first line, A = 5, B = 12, and C = -12.
For the second line, A = -1, B = 1, and C = -3.
Plugging these values into the formula, we get:
distance = |-12 - (-3)| / sqrt(5^2 + 12^2)
= |-12 + 3| / sqrt(25 + 144)
= |-9| / sqrt(169)
= 9 / sqrt(169)
= 9 / 13
So the distance between the lines 5x + 12y = 12 and y = -x + 3 is 9 / 13, which is approximately 0.692.
18. To find an equation of the line that bisects the acute angles formed by the lines 2x + y - 5 = 0 and 3x - 2y + 6 = 0, we need to find the slope of the bisecting line.
The slope of the bisecting line can be found using the formula:
m = (m1 + m2) / 2
where m1 is the slope of the first line and m2 is the slope of the second line.
For the first line, 2x + y - 5 = 0, we rewrite it in slope-intercept form as y = -2x + 5. So its slope is -2.
For the second line, 3x - 2y + 6 = 0, we rewrite it in slope-intercept form as y = (3/2)x + 3. So its slope is 3/2.
Plugging these values into the formula, we get:
m = (-2 + 3/2) / 2
= (-4/2 + 3/2) / 2
= -1/2 / 2
= -1/4
So the slope of the bisecting line is -1/4.
Since we know the slope and a point on the bisecting line (the point of intersection of the two given lines), we can use the point-slope form of a line to find its equation.
Let's say the point of intersection is (x0, y0). We can substitute this point and the slope (-1/4) into the point-slope form:
y - y0 = m(x - x0)
Plugging in the values, we get:
y - y0 = (-1/4)(x - x0)
So the equation of the line that bisects the acute angles is (-1/4)(x - x0) + y0.
19. To find an equation of the line that bisects the acute angles formed by the lines 4x + y + 3 = 0 and x + y - 2 = 0, we follow a similar process as in the previous question.
First, we find the slopes of the two given lines:
For the first line, 4x + y + 3 = 0, we rewrite it in slope-intercept form as y = -4x - 3. So its slope is -4.
For the second line, x + y - 2 = 0, we rewrite it in slope-intercept form as y = -x + 2. So its slope is -1.
Plugging these values into the formula, we get:
m = (-4 + (-1)) / 2
= -5/2
So the slope of the bisecting line is -5/2.
Similar to the previous question, we use the point-slope form of a line to find its equation. Let's say the point of intersection is (x0, y0). We can substitute this point and the slope (-5/2) into the point-slope form:
y - y0 = m(x - x0)
Plugging in the values, we get:
y - y0 = (-5/2)(x - x0)
So the equation of the line that bisects the acute angles is (-5/2)(x - x0) + y0.
20. If 90° < θ < 180°, we know that the cosine function is negative in this range. Since the cosine is negative, we can express it in terms of the tangent by using the trigonometric identity:
cos(θ) = -√(1 - tan^2(θ))
Therefore, the correct choice is -√(1 - tan^2(θ)).