I don't understand your question. Maybe it is just me.
Does someone in the forum understand this question?
If so, could they explain it to me please?
I am not able to solve this problem
Two fixed straight line OX and OY are cut by a variable line in the points A and B respectively and P and Q are feet of perpendiculars drawn from A and B upon the lines OBY and OAX. Show that if AB pass through a fixed point, then PQ will also pass through a fixed point.
What I am doing is
I am considering OX and OY as axes.
Thanx for advance
It yields to a co-ordinate geometry approach, I can't see anything better.
Let OX be the horizontal co-ordinate axis and let OY be the line with equation y=mx, (i.e., OY need not be the vertical co-ordinate axis).
Let the straight line AB, (A and B being the intercepts with OX and OY respectively) pass through the fixed point S(s,t) and have the variable gradient m'.
The equation of AB will be
Substituting y=0 gets us the co-ordinates of A,
Solving simultaneously y=mx with the equation of AB produces the co-ordinates of B,
For Q, drop a vertical from B onto OX, it will have co-ordinates
For P, drop a vertical from A onto OY. If P has co-ordinates (v,w), then since AP is perpendicular to OY, so
and since w=mv,
Having the co-ordinates of P and Q, the equation of the line PQ can be calculated,
It can now be verified that for any value of the variable gradient m', the line always passes through the point with