Question

Find the points where the line

$x=t+2,y=2t-5,z=-4t-7$

intersects each of the coordinate planes

Short answer

$\text{The required answer is}(0,-9,1)\left(\frac{9}{2},0,-17\right)\left(\frac{1}{4},\frac{-17}{2},0\right)\text{.}$

Step-by-step solution

Step 1:Given information

The line is

$x=t+2,y=2t-5,z=-4t-7$

Step 2:Calculation

$\text{Consider the line}\mathcal{L}\text{determined by the equations}x=t+2,y=2t-5,z=-4t-7\text{.}$

$\text{Take the line}\mathcal{L}\text{equation.}$

$x=t+2,y=2t-5,z=-4t-7$

Then,

$r\left(t\right)=(t+2,2t-5,-4t-7)$

When the line intersect in yz-plane the value of $x=0$

$\text{Now substitute}x=0\text{in}x=t+2\text{.}$

$0=t+2$

$\Rightarrow t=-2$

$\text{Substitute}t=-2\text{in}r\left(t\right)=(t+2,2t-5,-4t-7)\text{.}$

$r(-2)=(0,2·-2-5,-4·-2-7)$

$r(-2)=(0,-9,1)$

$\text{Thus the point in}yz\text{-plane is}(0,-9,1)$