What is the standard equation of the line perpendicular to line segment PQ at point Q? P(-3,-4) Q(7,-8)?

Question

The slope would be 5/2 right? because it is perpendicular.

but plugging that fraction into the point slope formula confuses me.
can anyone explain how to do this?

Answer 1

Point slope is y1-y2=M(x1-x2), so in your case it would be y+4=5/2(x+3) You add in this case because you have negative coordinates. Then distribute to get slope intercept form.

Answer 2

Okay, you are right about the slope of the perpendicular line: 5/2. Now you know everything about this new line except its y-intercept. Write what you have: Y = 5/2*X +B; next, substitute either point P or Q from above, lets choose P: -4=5/2*-3+B. Multiply the 5/2 by -3 to get -15/2. Now you have the following equation: -4= 15/2 +B, ( I know you hate fractions but there is no other way) now solve this for B. -4 -15/2=B
Simplify this to -8/2-15/2=b (common denominators), and now you are ready to solve. B should equal -23/2 or -12.5.

Answer 3

The slope of the line phase PQ is a million, so the perpendicular line has a slope of -a million. utilising y = mx+b replace the slope with -a million and use factor Q on your x and y values. then you would be waiting to remedy for b. with the help of how in case your answer must be in well-known variety, that variety is Ax + with the help of = C. So verify your answer is interior the spectacular format. wish this helps. good success.

Answer 4

Yes,STUDY and pay attention in class