Question

a line through point (3,0) meets the variable line y=tx at right angles at the point p.find in terms of t coordinate of p.​

Answer 1

Answer:

$(\frac{3}{t^{2}+1},\frac{3t}{t^{2}+1})$

Step-by-step explanation:

The variable line is y=tx .... (1)

So, It's slope is t. [ Similar to slope-intercept form of straight line, y=mx+c]

Now, the slope of straight line which is perpendicular to (1) will be -$(\frac{1}{t})$.

[ Since, the product of two straight line perpendicular to each other is -1]

Let us assume that the perpendicular straight line has the equation,

$y=-(\frac{1}{t})x+c$

⇒$y.t=-x+C$ { Where ct=C} ..... (2)

Now, it is given that equation(2) passes through (3,0) point.

So, putting x=3 and y=0 in equation (2), we get,

0=-3+C

⇒C=3

Hence, equation (2) becomes yt=-x+3 ..... (3)

Given condition is that equations (1) and (3) meets at p.

So, the co-ordinates of p will be obtained by solving (1) and (3).

So, we get,

$(tx)t=-x+3$

⇒ $x(t^{2}+1)=3$

⇒$x=\frac{3}{t^{2}+1}$ ..... (4)

Now, putting the value of x in equation (1), we get,

$y=tx=\frac{3t}{t^{2}+1}$..... (5)

Therefore, the co-ordinates of p will be $(\frac{3}{t^{2}+1},\frac{3t}{t^{2}+1})$

[From equations (4) and (5)]

(Answer)