Where is the triangle ?
Geometry
Level
2
If a triangle of area 9 square units is formed by the x-axis
and the lines x= 1 and y = mx - 4 (m < 0),
what are the coordinates of the vertices ?
(-8,0) (6,0) (1,8)
(-2,0) (1,0) (1,-6)
(-2,4) (1,5) (5,-6)
(2,3) (6,7) (9,8)
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1 solution
The x-axis is horizontal and the line x = 1 is vertical.
The third line with slope m < 0 goes from upper left to lower right, and intersects the y-axis at (0,-4). The line will intersect [x = 1] at (1,y) with y < -4, and the x-axis at (x,0) with x < 0.
Then the area of the triangle will be 1/2 (1-x) (-y) If that equals 9, then (1-x) (-y) = 18 or y (x-1) = 18.
Also, (y+4)/(1-0) = -4/(-x) = m ( slope from (x,0) to (0,-4) = slope from (0,-4) to (1,y) )
y+4 = 4/x (slope) xy + 4x = 4 xy - y = 18 (area) 4x + y = -14 (subtract) y = -4x - 14 Substituting back into the area equation, x(-4x-14) + 4x + 14 = 18 -4x^2 -14x + 4x - 4 = 0 -2x^2 - 5x -2 = 0 Roots are x = -2 and x = -1/2, with corresponding y values of -6 and -12.
So we have two possible triangles: (-2,0) (1,0) (1,-6) (area = 1/2 × 3 × 6) and (-1/2, 0) (1,0) (1,-12) (area = 1/2 × 1.5 × 12)