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maria [59]
3 years ago
9

Find the distance between the point (-2, -3) and the line with the equation y=4-2/3x

Mathematics
1 answer:
otez555 [7]3 years ago
4 0

Answer:

D = \frac{25\sqrt{13}}{13}

Step-by-step explanation:

Given

y = 4 - \frac{2}{3}x

(x_1,y_1) = (-2,-3)

Required

Determine the distance

y = 4 - \frac{2}{3}x

Write the above equation in standard form:

Ax + By + C = 0

So, we have:

\frac{2}{3}x+y  - 4 = 0

By comparison:

A = \frac{2}{3}  B = 1 and C = -4

The distance is calculated using:

D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

Where:

(x_1,y_1) = (-2,-3)

A = \frac{2}{3}  B = 1 and C = -4

This gives:

D = \frac{|\frac{2}{3} * (-2) + 1 * (-3) - 4|}{\sqrt{(\frac{2}{3})^2 + 1^2}}

D = \frac{|-\frac{4}{3} -3 - 4|}{\sqrt{\frac{4}{9} + 1}}

Take LCM

D = \frac{|\frac{-4-9-12}{3}|}{\sqrt{\frac{4+9}{9}}}

D = \frac{|\frac{-25}{3}|}{\sqrt{\frac{13}{9}}}

D = |\frac{-25}{3}|/\sqrt{\frac{13}{9}}

D = \frac{25}{3}/\sqrt{\frac{13}{9}}

Split the square root

D = \frac{25}{3}/\frac{\sqrt{13}}{\sqrt{9}}

Change / to *

D = \frac{25}{3}*\frac{\sqrt{9}}{\sqrt{13}}

D = \frac{25}{3}*\frac{3}{\sqrt{13}}

D = \frac{25}{\sqrt{13}}

Rationalize

D = \frac{25}{\sqrt{13}} * \frac{\sqrt{13}}{\sqrt{13}}

D = \frac{25\sqrt{13}}{13}

Hence, the distance is:

D = \frac{25\sqrt{13}}{13}

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