a straight line passes through p(-1,1) and q(3,4)

a) find the length of line pq
b) find the equation of the perpendicular bisector of the line pq leaving it in the form y=mx+c
c) determine the equation of line to line pq and passes through point (2,3) leaving your answer in double intercept form. hence state the y intercept

1 answer

(a) √((3+1)^2 + (4-1)^2) = √(4^2+3^2) = √25 = 5
(b) the midpoint of PQ is (P+Q)/2 = ((-1+3)/2,(1+4)/2) = (1,5/2)
The slope of PQ is (4-1)/(3+1) = 3/4
so the slope of the perpendicular is -4/3
So, the point-slope equation is y - 5/2 = -4/3 (x-1)
(c) Not sure what you mean by "line to line pq"
There are many lines that intersect pq. But once you figure out what that means, recall that if a line has
x-intercept = (a,0)
y-intercept = (0,b)
the intercept form of the line is
x/a + y/b = 1