Calculating the area of a minimum-area triangle . Two lines with equations: y = 2 – (1/2)x and (x-1) = 2 (y-1) respectively are graphed on a Cartesian plane.

Question

A line is drawn through the point (2, -1) which intersects the two previous lines to give a triangle of minimum area, with the point (2,-1) included in one side of that triangle. What is that area?

Answer 1

Transform the equations into standard form:

y = -(1/2)x + 2 and

y = (1/2)x + (1/2)

Since the point (2, -1) is on neither of these lines, it must therefore lie on the segment of the new line that comprises the 3rd side of the triangle.

It's fairly easy: these lines are R/L symmetric around the line x=(3/2), therefore the third side (as a line) has the form y= -1, and lies horizontal. Intersection points (vertices) are at (-3, -1) and at (6, -1). Area = (5/4)*(9/2) = 45/6.

Without that restriction "in one side of the triangle", the three lines could intersect at a point, triangle area = 0! Even a restriction "triangle is isosceles" won't give this desired triangle, but rather another: ((3/2,5/4),(2, 3/2),(2,1)).

Answer 2

I broke these points down. y=2-x/2

x=2y-1

I'm lost; the line from (2,-1) intersects the 2 previous lines at their intersection?

The triangle I get from the last line has pts (1.5,1) (1.75,0) (-1,0).

The 3rd line intersects x axis; beyond that point there's no other intersection.

Using the triangle that lies on x axis; I get 1.75*1/2=7/8 area.

Answer 3

Start by plotting the given lines. I do this with the DESMOS online tool: https://www.desmos.com/calculator

y = 2 – (1/2)x

(x-1) = 2 (y-1)

Locate the given point (2,-1) and visualize the various lines that can be drawn to form a triangle.

Now we can see an error in the problem:

"with the point (2,-1) included in one side of that triangle"---that clearly is impossible, so I think that point is supposed to be something else

Answer 4

Strategy: Convert to standard equation form, sketch the two initial lines, gauge the position of the required third side (part of a line), calculate the minimum triangle configuration.

Tactics: In standard form, the equations are: y = -(1/2) x + 2 and

y = (½) x + (½) . These cross at point ((3/2),(5/4)) . Since the slopes are opposite, calculating the slope of the third triangle side (containing the point) is simple: zero (an isosceles triangle will have the minimum area), with equation y = -1 . It's but a short calculation through to the x-values of the two lower vertices: 6 and -3 . So the triangle has base 9 and height (9/4), for area 8.125 .

Note that this problem was easier than the general case, in which the angle bisector has to be calculated from polar coordinate angles = arctan of each of the two initial lines, and the third side equation calculated from the perpendicular to that bisector (which also happens to be the other bisector of the original lines' intersection angle set) passing through the given point. These angles and slopes will in general not be rational numeric values, so you would have to approximate them.