+84 1)Find the equation whose graph is the line passing through the points (2, -3) and 94,5). Enter your answer in standard form Ax + By = C.
2)Find the area of the region that is enclosed by the graph of the equation 4x^2 - 12x + 4y^2 + 56y -471 = 0
3)The line 3x - 4y +24 intersects the x-axis and y-axis at points A and B respectively. Find distance AB.
4)Let be a point on the graph of equation, (x-94)^2 + (y-49)^2 = 44^2. What is the shortest possible distance between P and the x-axis?
5)The graph x^2 + y^2 = 100 intersects y = 3x at two points P and Q. Find PQ.
6)Given A = (1,1), B = (4,2), C = (3,4) and D = (2,3). Find the intersection of the diagonals of quadrilateral ABCD. Enter your answer in the form "(x,y)".
7)Let p and q be constants such that the graph of x^2 + y^2 - 6x +py + q is tangent to the y-axis. What is the area of the region enclosed by the graph?
8)In rectangle ABCD, E is the midpoint of BC and F is the midpoint of CD. Let G be the intersection of AE and BF. We know AB = 10 and BC = 15. Find DG.
Picture: https://latex.artofproblemsolving.com/1/7/5/1759cfad134302d78a2e2311360b02e6cc81fc65.png
9)The circles x^2 + y^2 = 4 and (x-2)^2 + (y-3)^2 = 7 intersect in two points A and B. Find the slope of AB.
10)A circle is tangent to the y-axis at the point (0,2) and passes through the point (8,0) as shown. Find the radius of the circle.
Picture:https://latex.artofproblemsolving.com/8/d/6/8d69e6ecf5ba0eefdc0b6b3d275445fc2fb678cc.png
11)The perimeter of square ABCD is 2. Two cars start from A and B simultaneously and drive clockwise on the perimeter of square ABCD in the same speed. A drone maintains its location as the midpoint of the two cars. How far did the drone fly after both cars get back to where they started?
+128564 1)Find the equation whose graph is the line passing through the points (2, -3) and (4,5). Enter your answer in standard form Ax + By = C.
Slope = [ 5 - -3 ] / [ 4 - 2 ] = 8/ 2 = 4
Equation is
y = 4 (x -4) + 5
y = 4x - 16 + 5
y = 4x - 11
4x - 1y = 11
+128564 2)Find the area of the region that is enclosed by the graph of the equation
4x^2 - 12x + 4y^2 + 56y -471 = 0 completre the square on x and y
4(x^2 - 3x + 9/4) + 4 (y + 14y + 49) = 471 + 9 + 196
4 ( x - 3/2)^2 + 4 ( y + 7)^2 = 676
(x - 3/2)^2 + (y + 7)^2 = 169
This is a circle centered at (3/2, -7) with a radius of 13
The area is pi * r^2 = 169 pi units^2
+128564 3)I'm assuming that we have.... 3x - 4y = 24 intersects the x-axis and y-axis at points A and B respectively. Find distance AB.
x intercept is (8,0)
y intercept is ( 0, -6)
The distance from A to B is
√ [ ( 8 - 0)^2 + ( -6 - 0)^2 ] =
√ [ 64 + 36 ] =
√100 =
10 units
+128564 5)The graph x^2 + y^2 = 100 intersects y = 3x at two points P and Q. Find PQ.
Sub the second function into the first function for y
x^2 + ( 3x)^2 = 100
x^2 + 9x^2 =100
10x^2 = 100 divide both sides
x^2 = 10 take both roots
x = √10 or x = -√10
So....let P = (√10, 3√10) and Q = ( -√10 , -3√10 )
So
PQ = √ [ (2√10)^2 + (6√10)^2 ] = √ ( 40 + 360) = √400 = 20 units
+128564 6)Given A = (1,1), B = (4,2), C = (3,4) and D = (2,3). Find the intersection of the diagonals of quadrilateral ABCD. Enter your answer in the form "(x,y)".
Probably several ways to do this, but let's get the equation of two lines and find their intersection
The line containing AC has a slope of [ 4 - 1 ] / [ 3 - 1 ] = 3/2
And the equation of this line is
y = (3/2) ( x - 1) + 1
y = (3/2)x +=- 1/2
The line containing BD has a slope of [ 3 - 2] / [ 2 - 4] = -1/2
And the equation of this line is
y = (-1/2)(x - 2) + 3
y = (-1/2)x + 4
To find the x intersection of these lines, we have
(3/2)x - 1/2 = (-1/2)x + 4
2x = 9/2
x = 9/4
And the y coordinate of the intersection is
y = (3/2)(9/4) - 1/2
y = 27/8 - 1/2
y= 27/8 - 4/8
y = 23/8
So...the intersection of the diagonals is ( 9/4, 23/8)
+128564 8)In rectangle ABCD, E is the midpoint of BC and F is the midpoint of CD. Let G be the intersection of AE and BF. We know AB = 10 and BC = 15. Find DG.
Let A = (0,0) B = (0, 10) C = (15,10) D =(15,0) E = ( 7.5, 10) F = (15, 5)
The equation of the the line containing segment AE is
y = (4/3)x
The slope of segment BF is [5 - 10] / [15 - 0] = -5/15 = -1/3
And the equation of the line containing BE is
y = (-1/3)x + 10
So....the x coordinate of the intersection of these two lines can be found as
(4/3)x = (-1/3)x + 10
(4/3 + 1/3) x = 10
(5/3)x =10
x = 6
And the y coordinate of the intersection of these lines is
y = (4/3)(6) = 8
So....G = ( 6,8)
So DG = √ [ (15 - 6)^2 + (8 - 0)^2 ] = √[9^2 + 8^2 ] = √[81 + 64] = √145 units
+128564 10)A circle is tangent to the y-axis at the point (0,2) and passes through the point (8,0) as shown. Find the radius of the circle.
Let the center of the circle be ( a, 2)
So.....the distance from the center of the circle to both points is the same....and we have
( a - 0)^2 + (2 - 2)^2 = ( 8 - a)^2 + (2 - 0)^2
a^2 = a^2 - 16a + 64 + 4 = 0
-16a + 68 = 0
-16a = -68
a = 68/16 = 17/4
So...the center is (17/4, 2)
And the radius is 17/4 = 4.25 units
+128564 9)The circles x^2 + y^2 = 4 and (x-2)^2 + (y-3)^2 = 7 intersect in two points A and B. Find the slope of AB.
Simplify the second equation
x^2 - 4x + 4 + y^2 - 6y + 9 = 7
x^2 -4x + y^2 - 6y = -6
Add both equations
x^2 - 4x + y^2 - 6y = - 6
x^2 + y^2 = 4
______________________
-4x - 6y = -10
Solve this for y
4x + 6y = 10
2x + 3y = 5
3y = 5 - 2x
y = [ 5 - 2x ] / 3 sub this into x^2 + y^2 = 4 for y
x^2 + ( [ 5 -2x ] / 3 )^2 = 4
x^2 + [ 4x^2 - 20x + 25 ] / 9 = 4
9x^2 + 4x^2 - 20x + 25 = 36
13x^2 - 20x - 11 = 0
Solving this equation for x produces
x = [10 + 9√3 ] / 13 or x = [ 10 - 9√3] / 13
And when x = [ 10 - 9√3] / 13, then y = [15 + 6√3 ] / 13
And when x = [10 + 9√3] / 13, then y = [15 - 6√3] / 13
So....the points of intersection are
( [ 10 - 9√3 ] / 13, [15 + 6√3 ] / 13 ) and ( [ 10 + 9√3 ] / 13, [15 - 6√3 ] / 13 )
And the slope between these points is
[ ( 15 + 6√3) - ( 15 - 6√3) ] / [ (10 - 9√3) - (10 + 9√3) ] =
[12√3 ] / [ (-18√3] =
-2/3
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