# Straight Line through a Fixed Point

• Dec 16th 2013, 06:20 PM
varunkanpur
Straight Line through a Fixed Point
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.

Attachment 29928
• Dec 16th 2013, 07:29 PM
Melody2
Re: Straight Line through a Fixed Point
Hi Varunkanpur,
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?
• Dec 18th 2013, 07:35 AM
BobP
Re: Straight Line through a Fixed Point
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 $\displaystyle y=m'x+(t-m's).$

Substituting y=0 gets us the co-ordinates of A, $\displaystyle \left ( \frac{m's-t}{m'},0 \right ).$

Solving simultaneously y=mx with the equation of AB produces the co-ordinates of B,$\displaystyle \left(\frac{m's-t}{m'-m},\frac{m(m's-t)}{m'-m}\right).$

For Q, drop a vertical from B onto OX, it will have co-ordinates $\displaystyle \left(\frac{m's-t}{m'-m},0\right).$

For P, drop a vertical from A onto OY. If P has co-ordinates (v,w), then since AP is perpendicular to OY, $\displaystyle \frac{w-0}{v-(m's-t)/m'}=-\frac{1}{m},$ so

$\displaystyle mm'w=m's-t-m'v,$ and since w=mv,

$\displaystyle v=\frac{m's-t}{m'(m^{2}+1)}$ and $\displaystyle w=\frac{m(m's-t)}{m'(m^{2}+1)}.$

Having the co-ordinates of P and Q, the equation of the line PQ can be calculated,

$\displaystyle y=x\frac{m-m'}{mm'+1}+\frac{m's-t}{mm'+1}.$

It can now be verified that for any value of the variable gradient m', the line always passes through the point with
co-ordinates

$\displaystyle \left(\frac{s+mt}{m^{2}+1},\frac{ms-t}{m^{2}+1}\right).$
• Dec 18th 2013, 10:29 PM
varunkanpur
Re: Straight Line through a Fixed Point
I thank you from answering my question, BobP
It help me a lot. I was struck at this question.

Again Thanx