SOLUTION: The vertices of a triangle ae P(-1,-1) Q(2,1) R(3,-1). Find the equation of the altitude of triangle PQR drawn from Q. There is a meteor shower, one meteor appears at A(7,8)

Question 61684: The vertices of a triangle ae P(-1,-1) Q(2,1) R(3,-1). Find the equation of the altitude of triangle PQR drawn from Q.

There is a meteor shower, one meteor appears at A(7,8) and disappears at B(1,4).
calculate the gradient of the meteor's flight path
and the acute angle between the flight path and the horizon (OX) correct to the nearest degree.

A second meteor, on a parallel path, starts at c (6,2) and disappears below the horizon at D, Find the co-ordinates of D.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
The vertices of a triangle ae P(-1,-1) Q(2,1) R(3,-1). Find the equation of the altitude of triangle PQR drawn from Q.
LET QA BE THE ALTITUDE.
IT IS PERPENDICULAR TO PR
SLOPE OF PR=[-1-(-1)]/[3-(-1)]=0
HENCE SLOPE OF QA IS -1/0 THAT IS -INFINITY
IT PASSES THROUGH Q (2,1).HENCE EQN. F ALTITUDE IS
X=1
There is a meteor shower, one meteor appears at A(7,8) and disappears at B(1,4).
calculate the gradient of the meteor's flight path
and the acute angle between the flight path and the horizon (OX) correct to the nearest degree.
SLOPE OF AB = (4-8)/(1-7)=-4/-6=2/3 IS THE GRADIENT OF HE METEOR' PATH.
ACUTE ANGLE = ARC TAN(2/3)=33.7 OR 34 DEGREES.
A second meteor, on a parallel path, starts at c (6,2) and disappears below the horizon at D, Find the co-ordinates of D.
PARALLEL LINE TO ABOVE WILL HAVE SAME SLOPE AS ABOVE .
IF D IS (H,K) , THEN
(K-2)/(H-6)=2/3
3(K-2)=2(H-6)
SINCE D IS GIVEN TO BE BELOW HORIZON,ASSUMING K TO BE ZERO,WE CAN FIND H
-6=-2(H-6)
2H=12+6=18
H=9
D IS (9,0)
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