The direction vector of the line
L: x=1+t, y=4t, z=2-3t
is <1,4,-3>
which is also the required normal vector of the plane.

Since the plane passes through point (-5,9,10), the required plane is :
Π 1(x-(-5)+4(y-9)-3(z-10)=0
=>
Π x+4y-3z=1

3 0

3 years ago

Please help this is hard math

Rainbow [258]

Answer:

$y=-4x+2$

Step-by-step explanation:

First we find the slope of the line AB passing through the point (2,0)
line AB is perpendicular to the line $y=\frac{1}{4}x-3$ , this equation is slope intercept for, which is $y=mx+b$ so our m or slope is $\frac{1}{4}$
Since the slope of line AB is perpendicular to $m=\frac{1}{4}$ then we use the equation $slope of line1\times slopeofline2=-1$ , so what do we multiply $\frac{1}{4}$ by to get -1? -4 right, so that's the slope of line AB with the point (2,0)
Now line CD is parallel to line AB, that means it has the same slope as AB, so slope of CD is -4
So we make the equation $y-y_1=m(x-x_1)$ so m= -4 and our point is (-1,6)
$y-(6)=-4(x-(-1)\\y-6=-4x-4\\y=-4x-4+6\\y=-4x+2$ thats the equation of line side CD