# Find the equation of straight line having a gradient of -1/t and passing through the point(t^2,2t).(pls see details below). ?

## This line meets the line 2y+x=4+2${t}^{2}$ at the point P. Show that the x-coordinates of P is 2t(t-1) and find the y-coordinates of P. Find the locus of P as t varies.

Jun 25, 2018

See below

#### Explanation:

If you know the slope $m$ (also known as gradient) of a line, and one its points $\left({x}_{0} , {y}_{0}\right)$, the equation of the line is given by

$y - {y}_{0} = m \left(x - {x}_{0}\right)$

In your case, $m = - \frac{1}{t}$, and $\left({x}_{0} , {y}_{0}\right) = \left({t}^{2} , 2 t\right)$. So, the equation is

$y - 2 t = - \frac{1}{t} \left(x - {t}^{2}\right)$

we may rewrite this equation as

$y = - \frac{1}{t} x + 3 t$

The other line is $2 y + x = 4 + 2 {t}^{2}$, which we can write as

$y = 2 + {t}^{2} - \frac{x}{2}$

(I just solved for $y$ bringing everything else to the right and dividing by $2$).

To find the point where the lines meet, let's ask for both equation to be satisfied. Since we know the expression for $y$ for both lines, let's set them to be equal:

$- \frac{1}{t} x + 3 t = 2 + {t}^{2} - \frac{x}{2}$

Solving for $x$, we get:

$- \frac{1}{t} x + \frac{x}{2} = {t}^{2} - 3 t + 2$

$x \left(\frac{1}{2} - \frac{1}{t}\right) = {t}^{2} - 3 t + 2$

$x \left(\setminus \frac{t - 2}{2 t}\right) = {t}^{2} - 3 t + 2$

$x = \setminus \frac{2 t}{t - 2} \left({t}^{2} - 3 t + 2\right)$

The last thing we need to do is to observe that ${t}^{2} - 3 t + 2$ can be factored as

${t}^{2} - 3 t + 2 = \left(t - 1\right) \left(t - 2\right)$

and thus the expression becomes

$x = \setminus \frac{2 t}{\cancel{t - 2}} \left(t - 1\right) \cancel{\left(t - 2\right)} = 2 t \left(t - 1\right)$

as required.