the equation of two sides of a triangle PQR are y-3x+6=0 and 3y-4y-2=0. D is the foot of the perpendicular from P to QR and the coordinates of D are (0, 4). find the equation of QR.
please explain .. thanx
2007-12-19 17:20:40 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
sori, it is 3y-4x-2=0..
2007-12-19 17:37:18 · update #1
here is an idea
http://ca.answers.yahoo.com/question/index;_ylt=AhAAi8qsBYFkcs2AcEV_IPTBFQx.;_ylv=3?qid=20071219150906AAPt1KW&show=7#profile-info-B2XZqa6yaa
http://ca.answers.yahoo.com/question/index;_ylt=Am.VAoKy31OppIpK.QYZV3jBFQx.;_ylv=3?qid=20071219145852AAm6558&show=7#profile-info-B2XZqa6yaa
2007-12-19 17:44:45 · answer #1 · answered by JavaScript_Junkie 6 · 1⤊ 0⤋
the hard part is making sense of your question. are there no x's in your 2nd equation? if so, it's a horizontal line, y = -2. what's the name of the point where those 2 lines intersect? P? since the other side is y = 3x-6, then with y = -2 at P, -2 = 3x-6, 4 = 3x, x = 4/3, so P is (4/3, -2).
slope of PD then is -6/(4/3) = -18/4 = -9/2. QR is perpendicular to that with a slope of 2/9, and it passes through D (0,4), so the equation of the line through QR must be
y-4 = 2/9(x-0)
y = (2/9)x + 4
or in the form you started with,
9y = 2x + 36
9y - 2x - 36 = 0
You sure about that 2nd equation?
2007-12-19 17:35:56 · answer #2 · answered by Philo 7 · 0⤊ 0⤋
I assume your second equation is 3y - 4x - 2 = 0
Solving the two given equations, P = (4, 6)
Now, slope of PD = (6 - 4)/(4 - 0) = 1/2
Hence, slope of QR = - 2
Hence, its equation is
y - 4 = -2 (x - 0)
2x + y = 4.
2007-12-19 17:29:28 · answer #3 · answered by Madhukar 7 · 0⤊ 0⤋
y -- 3x + 6 = 0 is PQ
3y -- 4x -- 2 = 0 is PR
then P is (4, 6) and
QR is 2(y -- 4) + x = 0 Or x + 2y = 8
2007-12-19 17:29:18 · answer #4 · answered by sv 7 · 0⤊ 0⤋
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2016-11-23 16:47:15 · answer #5 · answered by ? 4 · 0⤊ 0⤋