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Equation of line: 3x-7y+4 = 0

Base of triangle: 4/3

Height of triangle: 4/7

Area of triangle: 0.5*4/3*4*7 = 8/21

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6y ago
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6y ago

It is 8/21 square units.

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Q: What is the area of a triangle formed by the line 3x -7y plus 4 equals 0 when it meets the x and y axes on the Cartesian plane?
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Is an equilateral triangle am isosceles triangle?

Yes. An equilateral triangle is a triangle where all sides are equal and all angles are equal, therefore it meets the criteria of an isosceles triangle (having at least two sides that are equal).


What is the area of the triangle formed by the line 3x -7y plus 4 equals 0 and the axes on the Cartesian plane showing work?

Suppose the line meets the x and y axes at A and B (respectively). Then at A, x = 0 =-> -7y + 4 = 0 so y = 4/7. Therefore A = (0, 4/7) => |OA| = 4/7, and at B, y = 0 => 3x + 4 = 0 so x = -4/3. Therefore B = (-4/3, 0) => |OB| = 4/3. AOB is a right angled triangle with the right angle at O. Therefore area AOB = 1/2*|OA|*|OB| = 1/2*4/7*4/3 = 8/21 square units.


What is the area of a triangle formed when the line 3x -7y plus 4 equals 0 meets the x and y axes?

If: 3x-7y+4 = 0 then 3x-7y = -4 The intercepts are at: x is at (-4/3, 0) and y is at (0, 4/7) Distance of x from the origin: 4/3 Distance of y from the origin: 4/7 Area of the triangle: 1/2*4/3*4/7 = 8/21 square units


Which segment is an altitude of ΔABC?

A triangle has three altitudes so you need to be more specific. In general terms, it is a line from a vertex that meets the opposite side at right angles.


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For two sides to be parallel they must not meet at a point (vertex). A triangle has only 3 sides. Let the vertices of a triangle be ABC; then the 3 sides of the triangle are AB, AC and BC. The sides AB and AC meet at vertex A, so they cannot be parallel. The sides AB and BC meet at vertex B, so they cannot be parallel. So Side AB is not parallel to AC (meets at vertex A) and not parallel to BC (meets at vertex B). The only possibility left is that side BC is parallel to side AC. But the sides BC and AC meet at vertex C so they cannot be parallel. Thus none of the sides of a triangle can be parallel.

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What is the area of the triangle formed by the line 3x -7y plus 4 equals 0 and the axes on the Cartesian plane showing work?

Suppose the line meets the x and y axes at A and B (respectively). Then at A, x = 0 =-> -7y + 4 = 0 so y = 4/7. Therefore A = (0, 4/7) => |OA| = 4/7, and at B, y = 0 => 3x + 4 = 0 so x = -4/3. Therefore B = (-4/3, 0) => |OB| = 4/3. AOB is a right angled triangle with the right angle at O. Therefore area AOB = 1/2*|OA|*|OB| = 1/2*4/7*4/3 = 8/21 square units.


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What is the area of a triangle formed when the line 3x -7y plus 4 equals 0 meets the x and y axes?

If: 3x-7y+4 = 0 then 3x-7y = -4 The intercepts are at: x is at (-4/3, 0) and y is at (0, 4/7) Distance of x from the origin: 4/3 Distance of y from the origin: 4/7 Area of the triangle: 1/2*4/3*4/7 = 8/21 square units


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The sediments are called silt or alluvium. The land formed from these sedimants where the river meets the ocean is a delta.