A line through through the point (3,0) meet the variable line y=tx at right angle at the point P.
Find in terms of t, the co-ordinates of P
P has coordinates (x,tx)
The line from (3,0) to P has slope -1/t
now, drop an altitude from P to the x-axis.
tx/(x-3) = -1/t
x = 3/(1+t^2)
P = (3/(1+t^2), 3t/(1+t^2))
To find the coordinates of point P, we need to find the equation of the line passing through (3,0) and perpendicular to the line y = tx.
Let's first find the slope of the line y = tx. The equation is in the form y = mx, where m is the slope. Comparing this equation with y = tx, we can conclude that the slope of the line is t.
Now, we know that perpendicular lines have slopes that are negative reciprocals of each other. So, the slope of the line passing through (3,0) and perpendicular to y = tx will be -1/t.
We can use the point-slope form of a line to find the equation of this line. The point-slope form of a line is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Plugging in the values, we have:
y - 0 = (-1/t)(x - 3)
Simplifying this equation, we get:
y = (-1/t)x + 3/t
Now, let's find the coordinates of point P. Since point P lies on the line y = tx, we can substitute the equation of the line into the equation of y = tx to find the x-coordinate of P.
Substituting y = (-1/t)x + 3/t into y = tx, we have:
(-1/t)x + 3/t = tx
Multiplying both sides by t, we get:
- x + 3 = t^2x
Rearranging the equation, we have:
(1 + t^2)x = 3
Dividing both sides by (1 + t^2), we get:
x = 3 / (1 + t^2)
Now, substitute this value of x back into the equation y = tx to find the y-coordinate of P:
y = t * (3 / (1 + t^2))
Therefore, the coordinates of point P in terms of t are:
P(x, y) = (3 / (1 + t^2), t * (3 / (1 + t^2)))
line 1: y=tx
line 2: y=-x/t +b
but you know point 3,0 is on line 2
0=-3/t + b or b=3/t
line2: y=-x/t +3/t
line1 y=tx
set them equal
tx=-x/t+3/t solve for x
then solve for y.