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4 years ago

Fine the exact distance between the line 6x-y=-3 and the point (6,2). Show your work. Explain your answer.

Mathematics

1 answer:

AVprozaik [17]4 years ago

6 0

Answer:

$\sqrt{37}$ units

Step-by-step explanation:

If we draw a perpendicular from point (6,2) on the line 6x - y = - 3, then we have to find the length of the perpendicular.

We know the formula of length of perpendicular from a point $(x_{1}, y_{1} )$ to the straight line ax + by + c = 0 is given by

$\frac{|ax_{1} + by_{1} +c |}{\sqrt{a^{2}+b^{2}}}$

Therefore, in our case the perpendicular distance is

$\frac{|6(6)-2+3|}{\sqrt{6^{2} +(-1)^{2} } } = \frac{37}{\sqrt{37} } = \sqrt{37}$ units. (Answer)

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Given ABC is a right triangle.

AC is the hypotenuse and BD is the altitude.

AB and BC are legs of the triangle ABC.

AC = 16 and DC = 5

<u>Leg rule of geometric mean theorem:</u>

$\frac{\text { hypotenuse }}{\text { leg }}=\frac{\text { leg }}{\text { part }}$

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Do cross multiplication.

$\Rightarrow 16\times 5 = x\times x$

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Taking square root on both sides.

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square and square roots get canceled, we get

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