I get 9 triangle with fewer than 9 lines.
Draw a square: ABCD (4 lines)
Draw the diagonals AC, BD (2 lines) which meet at X in the centre.
On a separate part of the page, draw triangle PQR (3 lines).
That is 4 + 2 + 3 = 9 lines.
The triangles are:
ABC, BCD, CDA, DAB,
AXB, BXC, CXD, DXA,
and PQR
9 triangles with 9 lines. Could have done 13 triangles with 7 lines by drawing a line from A to BC.
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A line containing the point 5 9 and -5 -5 is going from the bottom left to the top right. It is a positive slope.
Each square has four corners, and each triangle has three corners. 5 squares X 4 corners/square = 20 corners 3 triangles X 3 corners/triangle = 9 corners 20 + 9 = 29 corners total.
4 - 9 = -5 One the number line . ... -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,.... On the number line , star at '4' and move '9' places (subtraction) to the left, you end up on '-5'.
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You zero out the variables in turn. Y = 9(0) + 5 Y = 5 ---------- 9X + 5 = 0 9X = - 5 X = - 5/9 Can use these points to graph line. ( - 5/9, 5)