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what is the area of a triangle with vertices at $(-2,1)$, $(2,1)$, and …

Question

what is the area of a triangle with vertices at $(-2,1)$, $(2,1)$, and $(3,4)$? enter your answer in the box. units²

Explanation:

Step1: Calculate the base length

The two points \((-2,1)\) and \((2,1)\) have the same \(y -\)coordinate. The distance between them (base \(b\)) is calculated using the distance formula for points \((x_1,y_1)\) and \((x_2,y_2)\) with \(y_1 = y_2\): \(b=\vert x_2 - x_1\vert\). So, \(b=\vert2-(-2)\vert=\vert2 + 2\vert=4\).

Step2: Calculate the height

The height \(h\) is the vertical distance from the line \(y = 1\) (where the base lies) to the point \((3,4)\). Since the \(y -\)coordinate of the base - line is \(y = 1\) and the \(y -\)coordinate of the third point is \(y = 4\), \(h=\vert4 - 1\vert=3\).

Step3: Calculate the area of the triangle

The area formula for a triangle is \(A=\frac{1}{2}\times b\times h\). Substitute \(b = 4\) and \(h = 3\) into the formula: \(A=\frac{1}{2}\times4\times3\). First, \(4\times3 = 12\), then \(\frac{1}{2}\times12=6\).

Answer:

\(6\)