distance between a line and point

Wording of the problem is questionable but understood.

They mean find the minimum distance between the point and the line
 
rachelmaddie said:
I don’t entirely understand this question.View attachment 14800
Click to expand...

If you've not been given a formula, then you can reason your way to a solution. I would begin by plotting the line and the point:

fmh_0091.png

Now, the minimum distance \(d\) between the point and the line will lie along a line that is perpendicular to the given line and which passes though the given point:

fmh_0092.png

Can you find the equation of the line along which the dashed segment lies? Once you have this line, then you will want to find where the two lines intersect, and then use the distance formula to find the distance between the given point and the point of intersection.

What do you find?
 
MarkFL said:
If you've not been given a formula, then you can reason your way to a solution. I would begin by plotting the line and the point:

View attachment 14806

Now, the minimum distance \(d\) between the point and the line will lie along a line that is perpendicular to the given line and which passes though the given point:

View attachment 14807

Can you find the equation of the line along which the dashed segment lies? Once you have this line, then you will want to find where the two lines intersect, and then use the distance formula to find the distance between the given point and the point of intersection.

What do you find?
Click to expand...
I think that the formula should be used.
 
rachelmaddie said:
I think that the formula should be used.
Click to expand...
@rachelmaddie, I agree with that. But I find that your greatest weakness is basic mathematics skills.
Write the line in standard form: \(\displaystyle 6x-y+3=0\).
The formula I gave in reply #5 tells us that \(\displaystyle D=\dfrac{|(6)(6)+(-1)(2)+(3)|}{\sqrt{(6)^2+(-1)^2}} \)
 
rachelmaddie said:
I think that the formula should be used.
Click to expand...

You didn't say whether you had been given a formula or not. If you wish to develop such a formula, then:

In the \(xy\)-plane, we have a point [MATH]P_0\left(x_0,y_0\right)[/MATH] separated from a line [MATH]y=mx+b[/MATH] by some distance [MATH]d>0[/MATH].

First, we find that the line perpendicular to [MATH]y=mx+b[/MATH] and passing through [MATH]P_0[/MATH] is:

[MATH]y=-\frac{1}{m}\left(x-x_0\right)+y_0[/MATH]
Solving the resulting linear system we find the common point to both lines is:

[MATH]\left(\frac{x_0+m\left(y_0-b\right)}{m^2+1},\frac{m\left(x_0+my_0\right)+b}{m^2+1}\right)[/MATH]
Now, using the distance formula for [MATH]P_0[/MATH] to the above point, we find:

[MATH]d=\sqrt{\left(x_0-\frac{x_0+m\left(y_0-b\right)}{m^2+1}\right)^2+\left(y_0-\frac{m\left(x_0+my_0\right)+b}{m^2+1}\right)^2}[/MATH]
The reader should verify that this reduces to:

[MATH]d=\frac{\left|mx_0+b-y_0\right|}{\sqrt{m^2+1}}[/MATH]
 
MarkFL said:
You didn't say whether you had been given a formula or not. If you wish to develop such a formula, then:

In the \(xy\)-plane, we have a point [MATH]P_0\left(x_0,y_0\right)[/MATH] separated from a line [MATH]y=mx+b[/MATH] by some distance [MATH]d>0[/MATH].

First, we find that the line perpendicular to [MATH]y=mx+b[/MATH] and passing through [MATH]P_0[/MATH] is:

[MATH]y=-\frac{1}{m}\left(x-x_0\right)+y_0[/MATH]
Solving the resulting linear system we find the common point to both lines is:

[MATH]\left(\frac{x_0+m\left(y_0-b\right)}{m^2+1},\frac{m\left(x_0+my_0\right)+b}{m^2+1}\right)[/MATH]
Now, using the distance formula for [MATH]P_0[/MATH] to the above point, we find:

[MATH]d=\sqrt{\left(x_0-\frac{x_0+m\left(y_0-b\right)}{m^2+1}\right)^2+\left(y_0-\frac{m\left(x_0+my_0\right)+b}{m^2+1}\right)^2}[/MATH]
The reader should verify that this reduces to:

[MATH]d=\frac{\left|mx_0+b-y_0\right|}{\sqrt{m^2+1}}[/MATH]
Click to expand...
You didn’t substitute the values?
 
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