find the equation of the bisector of the pairs of acute angles formed by lines x-2y+1=0 and x+3y-3=0.
2007-07-18 01:43:07 · 6 answers · asked by yeah_boy 2 in Science & Mathematics ➔ Mathematics
First, you need to find the point of intersection of the two lines by solving the system of equations. You can do this by solving for x in terms of y in the first equation, plugging the value for x into the second equation and solving for y, and using the value for y to solve either of the two equations for x. There are other ways to solve a system; any valid approach will be fine for this problem.
Second, find the slope of each of the two lines. The easiest way to do this is probably the rewrite each equation in slope-intercept form, y = mx + b, where m is the slope.
Finally, use the slope-point form of the equation of a line to get the equation for the bisector, which will share the point of intersection and have a slope equal to the arithmetic mean of the slopes of the original two lines. The slope-point form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Once you have the equation, you can rewrite it in any form you want.
2007-07-18 01:46:41 · answer #1 · answered by DavidK93 7 · 2⤊ 0⤋
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find the equation of the bisector of the pairs of acute angles formed by lines x-2y+1=0 and x+3y-3=0.
2015-08-10 12:12:03 · answer #2 · answered by Nia 1 · 0⤊ 0⤋
1. I would first get some graph paper and graph the triangle so that I could see what I was doing. 2. I would then use my knowledge of linear equations to determine the equations of those lines. 3. Add the pertinent equations of the two lines representing the angles together and divide them in half. That will give you the equations of the bisectors of the angles.
2016-03-17 00:01:52 · answer #3 · answered by ? 4 · 0⤊ 0⤋
x-2y + 1 = 0
2y = x +1
y = x/2 + 1/2 . . . . slope = 1/2
x+3y -3 = 0
3y = -x +3
y = -x/3 +1 . . . . . . slope = -1/3
intersection point
x/2 + 1/2 = -x/3 + 1
x/2 + x/3 = 1 - 1/2
(3x + 2x) / 6 = 1/2
5x = 3
x = 3/5
y = - 1/5 + 1 = 4/5
slope of the angle bisector = (1/2 -1/3)/2 = 1/6
equation of bisector line
1/6 = (y-4/5) / (x - 3/5)
x - 3/5 = 6(y-4/5)
x - 3/5 = 6y - 24/5
5x -3 = 30y -24
5x - 30y + 21 = 0 . . . . . . answer
2007-07-18 02:19:34 · answer #4 · answered by CPUcate 6 · 3⤊ 4⤋
U just directly hav to use the formula,
(x-2y+1)/root(5)=(x+3y-3)/root(10)
2007-07-18 01:47:26 · answer #5 · answered by aviral17 3 · 3⤊ 3⤋
root(2)abs.( x-2y +1) = abs.(x+3x-3)
2007-07-18 02:07:18 · answer #6 · answered by mramahmedmram 3 · 1⤊ 4⤋