Question

find the area of δabc with vertices a(4,-3) b(9,-3) and c(10,11) the area of abc is blank square units.

Explanation:

Step1: Find the length of AB

Points A(4, -3) and B(9, -3) have the same y - coordinate. The distance formula for two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Since \(y_1 = y_2=-3\), the length of AB is \(|x_2 - x_1|=|9 - 4| = 5\).

Step2: Find the height from C to AB

The line AB is horizontal (since y - coordinates are equal). The height from point C(10, 11) to AB is the vertical distance between the y - coordinate of C and the y - coordinate of AB. The y - coordinate of AB is - 3, so the height \(h=|11-(-3)|=|11 + 3| = 14\).

Step3: Calculate the area of the triangle

The formula for the area of a triangle is \(A=\frac{1}{2}\times base\times height\). Here, the base is AB = 5 and the height is 14. So \(A=\frac{1}{2}\times5\times14\).
\(\frac{1}{2}\times5\times14 = 5\times7=35\).

Answer:

35