Straight line stuff

[thomas2554]

i kinda did try to find out stuff and ended up nowhere. Can anyone give me some insights or even guide me through this?
I found the distance between each point in Q1, but then I'm stuck there.
So:
Dpq = Root85
Dpr = 4Root5
Drq = 3Root5

But then I think that Q2 and Q3 are linked to Q1 so can't really figure them out.
like for Q2a its said 'the equation of line p' and I think it's linking back to Q1.

 

[Otis]

I found the distance between each point in Q1

Hi Thomas. Thanks for explaining your approach. Question #1 asks for the equation of a line. We can find that without using distances.

Are you familiar with the Slope Formula, for finding the slope of a line passing through two known points? If so, then begin by calculating the slope of line QR (the line passing through points Q and R). Please share your calculation.

Next, line PS is perpendicular to line QR. Such slopes have a specific relationship: they are negative reciprocals of each other. In other words, if the slope of line QR is m, then the slope of perpendicular line PS is -1/m.

At this stage, we will know both the coordinates of a point on line PS and the line's slope. When we know a point and a slope, we can use the Point-Slope Formula to write the line's equation.

Please ask specific questions about any parts you don't understand, or show us your work.

Cheers
[imath]\;[/imath]

 

[Otis]

We can find [the equation] without using distances

Oops, I suppose that's a misstatement because my approach does use distances. The slope (steepness) of a line is defined as a ratio of two distances. If we make segment QR the hypotenuse of a right triangle and label the 90º vertex point V, then the Slope Formula first calculates the distances RV ("rise") and QV ("run") and then takes the ratio rise/run (as shown in the post#2 link).


[imath]\;[/imath]