maths problem?

Question

O(0,0),A(4,6),B(8,3) are vertices of the triangle OAB.Find:
1)Te equation of the median OM.
2)The equation of the altitude.
3)The co-ordinates of the 4th vertex C of the parallelogram OABC.

Answer 1

1)Equation of median OM is 9x-12y=0( M=(6,9/2) Midpoint formula, slope of OM is 3/4 and point on OM is O(0,0) and M(6,9/2)-Take any)
2) Equation of altitude from O is 4x-3y=0( slope of seg AB=-3/4 hence slope of its altitude=4/3 O(0,0) alone is the point on altitude. Get the equation from point slope formula)
3) Similarly, get equation for other altitudes: Equation for altitude from A is 8x+3y=50 and equation for altitude from B is2x+3y=25
4) Equation of seg thro' 0 is 3x+4y=0---(1)(slope being -3/4 which is also slope of seg AB as both segments are parallel)
Equation of seg thro' B is 3x-2y=18---(2). Solving (1) & (2) simultaneously gives C equal to C(4,-3)
5) Other two points could be similarly obtained to give points (-4,3) and (12,9)

Answer 2

The coordinates of M are (6,4.5).
Thus the equation of OM is y = 4.5/6 x or y = 3x/4.

The altitude you want is the perpendicular from A to OB. If so,
let AH be perpendicular to OB at H.
Slope of OB is 3/8 so slope of AH must be - 8/3.
y= -8/3x + b. Since AH passes through (4,6) we have
6 = (-8/3)(4) +b --> b = 50/3, so y = -8x/3 +50/3 is equation

Find coordinates of C(x,y)
The line CB has a slope = 3/2 = (3-y)/(8-x) <-- Eq 1
The line OC has a slope = -3/4 = y/x <-- Eq 2
Solve EQ 2 for x getting x = -4y/3. Put this in Eq 1 getting:
From Eq 2 it appears y= -3 and x= 4 Putting thes values in Eq 1 verifies it.
So C(4,-3) is the answer.

Answer 3

1. Find the midpoint of AB. Write the equation of the line including O and that point.
((8+4)/2, (6+3)/2) = (6,4.5) that'd be y = (4.5/6)x + 0
y=(3/4)x.

2. Find the slope of AB, m. Take the negative reciprocal of that slope (-1/m), and use that with the point-slope equation, using point 0,0. or y=(-1/m) + 0

m=(3-6)/(8-4)
m=-3/4
-1/m = 4/3
y=(4/3)x

3. find the intersection of: the line perpendicular to AB through B and the line perpendicular to OA through O.
line ┴ AB, we already detemined, has slope 4/3. Now we need to find the y-intercept. y=(4/3)x+b. Substitute (8,3) in for x and y
3=(4/3)(8) + b
b = 3-32/3 = (9-32)/3 = -23/3
so y = (4/3) x -23/3

line ┴ OA has slope -1/m of (6-0)/(4-0) = -1/m of (3/2) = -2/3. It passes through O, the origin, so it's equation is y = (-2/3)x
Equate the two y's to find their intersection

(4/3)x -23/3 = -2/3 x
-23/3 = 2/3x
-23=2x
x=-23/2

Substitute to find y:
y = (2/3) (-23/2)
y=-23/3

So (-32/2, -23/3) is the point. This answer looks wrong, so maybe I messed up ?

Answer 4

1.
midpoint of AB= (((4+8)/2),((6+3)/2))=(6,4.5)
equation of OM : y=mx+c (say)
m=4.5/6=3/4
c=0 as it passes through origin
=> eqn is 4y=3x

2.
eqn of AB: y=mx+c(say)
m=(6-3)/(4-8)=-3/4
slope of altitude m'= 4/3
as it passes thourgh origin y-intercept=0
eqn. of altitude perp to AB => 3y=4x

3.
using the rule for cyclic notation of a parallelogram:
OA || CB
=> slope of OA = slope of BC= 6/4=3/2
if eqn. of BC is y=(3/2)x+c as it passes through B
c= (3/2*8) - 3 = 9....using BODMAS
thus if C's absissa is x then it's ordinate:
y=(3x+18)/2 = (3/2)(x+6)

length of AB=sqaure root of ((4)^2+(3)^2)=5
being a parallelogram implies that OC=AB

x^2+ [(3/2)(x+6)]^2=25
4x^2+9x^2+108x+324=100...simplifying the eqn.
13x^2+108x+224=0
too complicated...
so try correlating OA and BC.
and simplifying.
Sorry but this is where my mental maths gets overloaded.

Answer 5

1). m=( x1+x2)/2,(y1+y2)/2 = 12/2,9/2 = (6,4.6555) om=(3,2.3333) 3).. 1/2*((-4)(0)+(4)4+(8*3)) =40/2 =20