The equation y-5=m(x-2) represents a line, no matter what value m. Show that the first quadrant triangle is always at least 20.

Question

The equation y-5=m(x-2) represents a line, no matter what value m.

(A) what are the x and y intercepts of this line? I got: 2-5/m =x ; -2m+5=y

(B) for what value of m does this line form a triangle of area 36 with the positive axes? I got m= -1/2 or m=-25/2

(C) The equation y-5=m(x-2) represents a line, no matter what value m. Ans? I am stuck on this one. I multipled x*y=0 and got m=5/2 but i know the slope has to be negative since it's in quadrant one. If i use m=-5/2 and solve for x, y (using answers from a), and then A=(xy)/2, I get 20, but not sure i can just make m negative? (this is an 8th grade question- so pre-calc).

Answer 1

You got A and B right. Perfect

No matter what m is? If m = 0, the line is y= 5

If m= negative or positive infinity, the line is x = 2

in some sense, that gives an infinite area, definitely > 20

but m>0 there is no triangle, m has to be negative

if m= -5/2, then

y=(-5/2)(x-2) + 5

y intercept is 10

x intercept = 4

Area = 10(4)/2 = 20

Looks good.

Is this in a text book problem or an instructor problem. Can't believe they do this in the 8th grade. Seems too unnecessary.

Take the given point (2,5), double each of the x and y coordinates to get 4,10. That gives the x and y coordinates that create the minimum area triangle. -10/4 = -5/2 = m, the slope of the line through that point with minimum area = 20. (0,10) and (4,0) are the y and x intercepts the create the smallest area with a line through (2,5). Area = base of 4 times height of 10 divided by 2. A=bh/2