How to find the value of k in (k-5)x+(4k^2-9)y+(6k^2-7k-3)=0 ?

[AlphaCentauri]

How to find the value of k from the linear equation (k-5)x+(4k²-9)y+(6k²-7k-3)=0 given that the line is parallel to y-axis?

 

[Deleted member 4993]

How to find the value of k from the linear equation (k-5)x+(4k²-9)y+(6k²-7k-3)=0 given that the line is parallel to y-axis?

What is the general equation of a line parallel to y-axis?

 

[AlphaCentauri]

What is the general equation of a line parallel to y-axis?

From what I know, it's x=a because y = 0
When I tried to apply that into this equation, I get the equation (k-5)x+(6k²-7k-3)=0 but I don't know what to do next.

 

[Dr.Peterson]

From what I know, it's x=a because y = 0
When I tried to apply that into this equation, I get the equation (k-5)x+(6k²-7k-3)=0 but I don't know what to do next.

You can't just drop the y term; you have to make it zero by choosing the right value of k ...

 

[pka]

How to find the value of k from the linear equation (k-5)x+(4k²-9)y+(6k²-7k-3)=0 given that the line is parallel to y-axis?

In general \(\displaystyle Ax+By+C=0\) is a line.
If \(\displaystyle A\cdot B\ne 0\) then this line is not parallel to either the \(\displaystyle x-\text{axis}\) nor the \(\displaystyle y-\text{axis}\).
If \(\displaystyle A=0~\&~B\ne 0\) then the line is parallel to the \(\displaystyle x-\text{axis}\) and if \(\displaystyle B=0~\&~A\ne 0\) then the line is parallel to the \(\displaystyle y-\text{axis}\)

Now for what value of \(\displaystyle k\) is the given line parallel to the \(\displaystyle y-\text{axis}\)

 

[Harry_the_cat]

You can't just drop the y term; you have to make it zero by choosing the right value of k ...

I don't think AC just dropped the y-term … I think he put in y=0. Why though? y=0 is the x-axis … makes no sense to sub y=0.

 

[Harry_the_cat]

From what I know, it's x=a because y = 0
When I tried to apply that into this equation, I get the equation (k-5)x+(6k²-7k-3)=0 but I don't know what to do next.

Yes the general equation of a line parallel to the y-axis has the form x=a ie x = a constant, ie x = 0y +a (Note this doesn't mean that y=0, it means that the coefficient of y is 0)
Try this:
1. Rearrange the equation to make x the subject.
2. Let the coefficient of y be 0
3. Solve for k

 

[HallsofIvy]

Your original equation was \(\displaystyle (k- 5)x+ (4k^2- 9)y+ (6k^2- 7k- 3)\)= 0.

NO, you do not set y= 0. This is to be an equation that is true for all y. Instead, since the equation of a line parallel to the y-axis is x= a with no "y", what value of k makes the coefficient of y equal to 0?

 

[AlphaCentauri]

Okay, now I understand. If \(\displaystyle 4k^2-9=0\), then \(\displaystyle k=\pm\dfrac{3}{2}\) right?

Thanks for your help.