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# The length of altitude through A of the triangle ABC where A = (-3, 0), B = (4, -1), C = (5, 2) is:

 A) $$\Large \frac{2}{\sqrt{10}}$$ B) $$\Large \frac{4}{\sqrt{10}}$$ C) $$\Large \frac{11}{\sqrt{10}}$$ D) $$\Large \frac{22}{\sqrt{10}}$$

 D) $$\Large \frac{22}{\sqrt{10}}$$

In $$\Large \triangle ABC$$ the vertices are $$\Large A \left(-3,\ 0\right),\ B \left(4,\ -1\right)\ and\ C \left(5,\ 2\right)$$

Distance between B and C

$$\Large BC=\sqrt{ \left(5-4\right)^{2}+ \left(2+1\right)^{2} } = \sqrt{1+9} = \sqrt{10}$$

Area of $$\Large \triangle ABC$$

= $$\Large \frac{1}{2} \left[ x_{1} \left( y_{2} - y_{3} \right) + x_{2} \left( y_{3} - y_{1} \right) + x_{3} \left( y_{1} - y_{2} \right) \right]$$

= $$\Large \frac{1}{2}\left[ -3 \left(-1-2\right)+4 \left(2-0\right)+5 \left(0+1\right) \right]$$

= $$\Large \frac{1}{2}\left[ 9+8+5 \right]=11$$

As we know area of $$\Large \triangle ABC = \frac{1}{2} \times BC \times AL$$

=> $$\Large 11 = \frac{1}{2} \times \sqrt{10} \times AL$$

=> $$\Large AL = \frac{2 \times 11}{\sqrt{10}}=\frac{22}{\sqrt{10}}$$

Part of solved Rectangular and Cartesian products questions and answers : >> Elementary Mathematics >> Rectangular and Cartesian products

Similar Questions
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