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Akimi4 [234]
3 years ago
8

(-9z - 8) - (4z - 7)

Mathematics
1 answer:
inn [45]3 years ago
7 0
The answer is -13z - 1
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<h2>The area of a triangle is =54 square units</h2><h2>The perpendicular distance from B to AC is = \frac{108}{\sqrt{149} } units</h2>

Step-by-step explanation:

Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

x_1=2,y_1=1,x_2=12,y_2=2,x_3=12 and      y_3=8

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The length of AC = \sqrt{(x_1-x_3)^{2} +(y_1-y_3)^2}

                          = \sqrt{(2-12)^{2} +(1-8)^2}

                         =\sqrt{149} units

Let the perpendicular distance from B to AC be = x

According To Problem

\frac{1}{2} \times  x \times \sqrt{149} = 54

⇔x =\frac{108}{\sqrt{149} } units

Therefore the perpendicular distance from B to AC is = \frac{108}{\sqrt{149} } units

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