+4 1. The circle (x + 3)2 + (y − 4)2 = 36 and the line 3x − 4y + 10 = 0 intersect twice as shown below. What is the distance from the center of the circle to the line?
2. A line goes through (2, 4) and (−8,−1). The line has its x-intercept at point P and its y-intercept at point Q. Find the area of △OPQ, where O is the origin.
3.In △ABC, AB = 4 and BC = 5. Find the number of possible integer values of CA. Reminder: In this class, we do not consider degenerate triangles (i.e. a triangle whose area is 0).
4.In obtuse △ABC, AB = 4 and BC = 5. Find the number of possible integer values of CA.
5. A, B, C, and D are points on a circle, and segments AC and BD intersect at a point P inside the circle, such that AP = 8, PC = 9, and BD = 22. Find BP, given that BP < DP.
6.] In △ABC, sin A : sin B : sinC = 4 : 5 : 6. Find cos(A + C).
THank youuu
+1768 Problem 2
First we will find the slopes of the lines that go through P and Q respectively. The slope of the first line is xP−xOyP−yO=4. The slope of the second line is xQ−xOyQ−yO=−1. We know the angle between the lines is 180 degrees because they are perpendicular to each other. Then, the slope of the first line is −1 times the reciprocal of the slope of the second line which means the slope is −11 which is −1. Now we can find the equations of the lines. The equation of the line that goes through P is y−4=−1(x−2) and the equation of the line that goes through Q is y−0=−1(x+8). We can find the coordinates of P by letting y=0 in the first equation. We get 0−4=−1(x−2) and x=6. We can find the coordinates of Q by letting x=0 in the second equation. We get y−0=−1(0+8) and y=−8. The area of triangle OPQ=8*6/2=24.
+1768 Problem 3
By the triangle inequality, (5-4 \leq CA \leq 5+4), so 1≤CA≤9. Therefore, there are 9 possible integer values of CA.
Problem 4
By the Triangle Inequality, we know that 3 <9. We also know that the area of an obtuse triangle is positive, so CA=4 and CA=5. Therefore, the only possible integer values of CA are 3,6,7,8.
