the equation of two sides of a triangle PQR are y-3x+6=0 and 3y-4x-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 alot
the answer is 2x+y-4=0.. but please show me how u get it.. thanx
2007-12-20 11:15:32 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
P is the intercept of the given sides as D does not belong to any of them
y= 3x-6 and 9x-18-4x-2=0 so 5x = 20 x= 4 and y= 6 so
P(4,6)
The slope of PD is (6-4)/(4-0)= 1/2 so the slope of QR is -2
so QR) y-4=-2x so 2x+y-4=0
2007-12-20 11:52:56 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
the confusing section is making experience of your question. are there no x's on your 2d equation? if so, it is a horizontal line, y = -2. what's the call of the factor the place those 2 lines intersect? P? considering the fact that the different component 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 two/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 interior the type you started with, 9y = 2x + 36 9y - 2x - 36 = 0 You constructive approximately that 2d equation?
2016-12-11 10:29:37 · answer #2 · answered by Anonymous · 0⤊ 0⤋