Help with task

That's the way! Next task is to find a line perpendicular to [MATH]y = 2x + 3[/MATH] that also passes through the point (1, 1).

Once again, you can find a perpendicular slope by taking [MATH]-\frac{1}{m}[/MATH]. This is something to be memorized, as you don't want to have to look at a graph every time.

The perpendicular line will also need a different y-intercept, as it passes through (2, 2) if you leave the y-intercept at 3.
 
swag312 said:
Yeah I understand it now, y and x has to be 1 and 1, so you have to modify b in order for it to happen which leads to y=2x-1
Click to expand...

[MATH]y=m(x-x_1)+y_1=2(x-1)+1=2x-2+1=2x-1\quad\checkmark[/MATH]
So, what is the slope for the perpendicular line?
 
swag312 said:
I'm not sure but maybe it's y = [MATH]-\frac{1}{2x}+1.5[/MATH] ?
Click to expand...

To find the slope of the perpendicular line:

[MATH]2m=-1\implies m=-\frac{1}{2}[/MATH]
Hence:

[MATH]y=m(x-x_1)+y_1=-\frac{1}{2}(x-1)+1=-\frac{1}{2}x+\frac{3}{2}[/MATH]
It looks like you may have been correct, except for a typo putting \(x\) in the denominator.
 
swag312 said:
Write down for the line y = 2x + 3
perpendicular and parallel lines passing through the point (1; 1) equations.
Click to expand...
Your line is standard form is \(\displaystyle 2x-y+3=0\)
Any line parallel to that line is \(\displaystyle 2x-y+\large{?}=0\) all one needs is to use the point (1,1) to find the ?
Any line perpendicular to the given line is \(\displaystyle x+2y+{\large{?}}=0\) all one needs is to use the point (1,1) to find the ?
 
swag312 said:
Yeah I understand it now, y and x has to be 1 and 1, so you have to modify b in order for it to happen which leads to y=2x-1
Click to expand...
OK, very good.
Now can you try to find the equation of the perpendicular line? Remember that if m=2 for the original line than the perpendicular line will have a slope of -1/2. It will also pass through the point (1,1).
 
pka said:
Your line is standard form is \(\displaystyle 2x-y+3=0\)
Any line parallel to that line is \(\displaystyle 2x-y+\large{?}=0\) all one needs is to use the point (1,1) to find the ?
Any line perpendicular to the given line is \(\displaystyle x+2y+{\large{?}}=0\) all one needs is to use the point (1,1) to find the ?
Click to expand...
This is a mini-essay on linear standard form.
Consider a given line \(\displaystyle \ell_0: Ax+By+C=0\) as a standard form.
If \(\displaystyle A\cdot B\ne 0\) then the line is neither vertical or horizontal.
In that case we can read off from the standard form the line's slope: \(\displaystyle \frac{-A}{B}\).
Any line parallel to \(\displaystyle \ell_0\) has the form \(\displaystyle \ell_{\|}: Ax+By+D=0\)
Moreover, if \(\displaystyle (p,q)\in\ell_{\|}\) then \(\displaystyle D=-(Ap+Bq)\)

Any line perpendicular to \(\displaystyle \ell_0\) has the form \(\displaystyle \ell_{\bot}: Bx-Ay+E=0\)
Moreover, if \(\displaystyle (p,q)\in\ell_{\bot}\) then \(\displaystyle E=-(Bp-Aq)\).

Here is just an interesting aside: If \(\displaystyle h\cdot k\ne 0\) then \(\displaystyle \frac{x}{h}+\frac{y}{k}=1\)
That is also a line with \(\displaystyle x\text{-intercept} (h,0)~\&~y\text{-intercept} (0,k)\)

 
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