Within an orthonormal system consider points A(1,4) B(5,1) C(4,8)
(a)Calculate the angle ABC
(b)Show that 3x+4y-19=0 is an equation for line L which passes through A and B
(c)Find a vector equation for line L
(d)(i)give an equation for the circle with the line segment AB as diameter
(ii)Give an equation for the tangent to the circle at the point B
(e)(i)Find the distance from O,the origin,to line L
(ii)Give an equation for the line through O and perpendicular to L
(iii)Find an equation for the smallest circle through circle O which has a segment of L as a diameter
(f)Calculate the area of the triangle
ABC
I need mostly need help on c,d,e
Vector BA = -4i + 3j
Vector BC = -1i + 7j
|BA| = sqrt (16+9) = 5
|BC| = sqrt (1 + 49) = sqrt (50)=5sqrt2
BA dot BC = 4+21 = 25
BA dot BC = |BA| |BC| cos ABC
so
25 = 5 * 5 sqrt 2 * cos ABC
cos ABC = sqrt 2 / 3 = .707
ABC = 45 degrees
b
L is y = -(3/4) x + 19/4
note slope of vector AB = -3/4
now does L go through A which is (1,4)?
4 = -(3/4)x + 19/4
16 = -3x + 19
x = 1 sure enough L goes through A and B
c
did that right off in A but I did BA = -4i + 3 j
we want AB = 4i -3j = -BA
d
center of circle at center of AB
x = (5+1)/2 = 3
y = (4+1)/2 = 2.5
radius = |BA|/2 = 2.5
so
(x-3)^2 +(y-2.5)^2 = 2.5^2 = 6.25
tangent at B has slope perpendicular to L -1/-(3/4) or 4/3 and goes through B
y = (4/3) x + b
1 = (4/3)5 + b
b = 1 - 20/3 = -17/3
so y = (4/3) x -17/3
or 3 y = 4 x - 17
e
call line perpendicular to L through the origin P
equation of P is
y = m' x + 0 (because through origin b =0)
m' = -1/slope of L = 4/3
so
y = (4/3)x is equation of P (that is e ii by the way
where does L hit P ?
(4/3) x = -(3/4) x + 19/4
(4/3 + 3/4) x = 19/4
(16 + 9 ) x/12 = 19/4
25 x = 3*19
x = 3*19/25
y = 4*19/25
the distance from O to that intersection is (19/25)sqrt (3^2+4^2) =(19/25)*5 = 19/5 that is e i answer
e iii
well I know center is at x = 3*19/25 and y = 4*19/25 and the radius is 19/5
so
(x - 3*19/25)^2 + (y - 4*19/25)^2 = 19^2/25^2
f
area = (1/2) |BA||BC| sin ABC
= (1/2)(5)(5 sqrt 2)(sin45)
= (25/2) sqrt 2 * 1/sqrt 2
= 25/2 = 12.5
(c) To find a vector equation for line L, we can use the fact that any point on the line can be written as the sum of a fixed point on the line (let's call it P) and a scalar multiple of the direction vector of the line (let's call it d).
We need to find P and d. We are given that line L passes through points A(1,4) and B(5,1). Hence, P can be any of these points. Let's choose P = A(1,4).
Next, we need to find the direction vector d. We can find the direction vector by taking the difference between any two points on the line. Let's take the difference between points A and B:
d = B - A = (5,1) - (1,4) = (4,-3).
Now, we can write the vector equation for line L as:
r = P + td,
where r is any point on line L, t is a scalar parameter, P is the fixed point on the line (in this case, A), and d is the direction vector (in this case, (4,-3)).
So, the vector equation for line L is:
r = (1,4) + t(4,-3), where t is a scalar.
(d)
(i) To find the equation for the circle with the line segment AB as diameter, we can start by finding the midpoint M of AB. The midpoint formula is:
M = ( (x₁ + x₂)/2, (y₁ + y₂)/2 ),
where (x₁, y₁) are the coordinates of point A and (x₂, y₂) are the coordinates of point B.
Using the given coordinates of points A(1,4) and B(5,1), we can find the midpoint M:
M = ( (1+5)/2, (4+1)/2 ) = (3, 2.5).
The coordinates of the midpoint M are (3, 2.5). This point will be the center of the circle.
The radius of the circle can be found by finding the distance between points A and B, and then dividing by 2:
radius = AB/2 = √((x₂ - x₁)² + (y₂ - y₁)²)/2.
Using the coordinates of points A(1,4) and B(5,1), we can find the radius:
radius = √((5-1)² + (1-4)²)/2 = √(16 + 9)/2 = √25/2 = 5/√2.
So, the equation for the circle with line segment AB as diameter is:
(x-3)² + (y-2.5)² = (5/√2)².
(ii) To find the equation for the tangent to the circle at the point B, we need to find the slope of the tangent line. The slope of a line tangent to a circle at a given point is perpendicular to the radius drawn to that point.
We already know the center of the circle is (3, 2.5), and the radius is 5/√2. The coordinates of point B are (5, 1).
The slope of the radius drawn from the center to point B is given by the difference in y-coordinates divided by the difference in x-coordinates:
slope of the radius = (1 - 2.5)/(5 - 3) = -1.5/2 = -0.75.
Since the tangent line is perpendicular to the radius, the slope of the tangent line will be the negative reciprocal of the slope of the radius.
slope of the tangent = -1/(slope of the radius) = -1/(-0.75) = 4/3.
Now we have the slope of the tangent line. We can use the point-slope form of a line to write the equation of the tangent line. We know that point B(5, 1) lies on the tangent line.
y - y₁ = m(x - x₁),
where m is the slope of the line and (x₁, y₁) are the coordinates of the point on the line.
Using the slope m = 4/3 and the coordinates (x₁, y₁) = (5, 1), we can write the equation of the tangent line:
y - 1 = (4/3)(x - 5).