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The points A(-3,3), B(2,-1) and C(5,1) are given and O is the origin. Find in the form ax+by+c=0 and equation of the line:

a) l1 passing through A and C
b) l2 through B perpendicular to AC
c) l3 through M, the midpoint of AB, and parallel to AC.

Working Out Would Be Nice, Im Really Stuck :( Thanks In Advance.

2007-03-29 06:25:43 · 5 answers · asked by kingbeer00 1 in Science & Mathematics Mathematics

Btw - Another Addication Question

d) Find the Co-ordinates of the point where the line l3 and BC meet.

2007-03-29 08:21:44 · update #1

5 answers

a)y=mx+c
At (-3,3),
3 = -3m + c ( eqn 1)
At (5,1)
1 =5m + c ( eqn 2)
(eqn 2) - (eqn 1),
-2 = 8m
m = -1/4
From ( eqn 1)
3 = (-1/4)(-3)+c
c =3 -3/4
c = 9/4
Therefore y = (-1/4)x+(9/4)
4y = -x +9
Answer: x+4y-9 =0

b) From a), we can see that gradient of AC = (-1/4)
Using the fact that m1xm2 = -1
m2 = -1/(-1/4)
m2 = 4

Using y= mx+c
Since m2 = 4
Therefore y = 4x+c
At B(2,-1)
-1 = 4(2)+c
c =-9
Therefore y = 4x-9
Answer: 4x-y-9=0

c) M = [(-3+2)/2,(3-1)/2)
M = (-1/2,1)
Using y = mx+c
Since l3 is parallel to AC, gradient = -1/4
y = (-1/4)x+c
At M(-1/2,1)
1 = (-1/4)(-1/2)+c
c = 1+1/8
c = 9/8
Therefore y = (-1/4)x+9/8
Answer: 2x+8y-9=0

Note: I just mentally do only. Accuracy of figure you check yourself. Important thing is to stick to the concepts.

2007-03-29 06:59:22 · answer #1 · answered by Michael Toh 3 · 0 0

1) Passing through A and C
equation of line y = mx+c
slope = (y2-y1)/(x2-x1) = -2/8
y = -2x/8 +c
plugging in values (-3,3) 3 = 6/8+c => c = 9/4

y = (-x+9)/4
or x + 4y -9 = 0

b) slope of a line perpendicular to AC is 4
condition is m1m2 = -1
y = 4x +c
plugging in (2,-1) => c = -9
or y-4x+9 = 0
Answer -4x+y + 9 = 0

c) midpoint of AB = ((x1+x2)/2,(y1+y2)/2) = (-1/2,1)
its parallel to AC , so slope is same , slope = -1/4
y = -x/4 +c

plugging in (-1/2,1) => 1 = 1/8 +c => c = 7/8
8y = -2x +7
or -2x + 8y -7 = 0

Answer -2x + 8y -7 = 0

2007-03-29 13:50:57 · answer #2 · answered by Nishit V 3 · 0 0

1) Passing through A and C
equation of line y = mx+c
slope = (y2-y1)/(x2-x1) = -2/8
y = -2x/8 +c
plugging in values (-3,3) 3 = 6/8+c => c = 9/4

y = (-x+9)/4
or x + 4y -9 = 0

b) slope of a line perpendicular to AC is 4
condition is m1m2 = -1
y = 4x +c
plugging in (2,-1) => c = -9
or y-4x+9 = 0
Answer -4x+y + 9 = 0

c) midpoint of AB = ((x1+x2)/2,(y1+y2)/2) = (-1/2,1)
its parallel to AC , so slope is same , slope = -1/4
y = -x/4 +c

plugging in (-1/2,1) => 1 = 1/8 +c => c = 7/8
8y = -2x +7
or -2x + 8y -7 = 0

Answer -2x + 8y -7 = 0

2007-04-05 17:05:01 · answer #3 · answered by cash 2 · 0 0

Question a)
A(-3,3) and C(5,1)
m AC = - 2/8 = - 1/4
y - 1 = (-1/4).(x - 5)
4y - 4 = - x + 5
x + 4y - 9 = 0

Question b)
mAC = - 1/4, B(2, -1)
perpendicular has m = 4
y + 1 = 4(x - 2)
y + 1 = 4x - 8
4x - y - 9 = 0

Question c)
mid point of AB is (-1/2,1) and mAC = -1/4
y - 1 = -(1/4).(x + 1/2)
4y - 4 = - x - 1/2
8y - 8 = - 2x - 1
2x + 8y - 7 = 0

2007-03-29 14:18:47 · answer #4 · answered by Como 7 · 0 0

Hi,
pl. contact me for solutions.
I have master degree in mathematics and I have got thirteen ten years of teaching experience in Math at college level and currently I am working as a lecturer for an army engineering college author of math guide.

thankyou

2007-04-03 05:25:15 · answer #5 · answered by valivety v 3 · 0 1

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