Answer to Question #252729 in Analytic Geometry for ling

Question #252729

5. Find the angle from the line 2x + y – 8 = 0 to the line x + 3y + 4 = 0. Draw the lines and

locate the angle formed.

6. Two sides of a square lie along the lines 2y = 20 – 3x and 3x + 2y = 48. Find the area of

the square.


1
Expert's answer
2021-10-19T13:46:40-0400

5.


"2x + y - 8 = 0=>y=-2x+8"

"slope_1=m_1=-2"

"x + 3y + 4 = 0=>y=-\\dfrac{1}{3}x-\\dfrac{4}{3}"

"slope_2=m_2=-\\dfrac{1}{3}"

"\\theta=\\tan^{-1}\\dfrac{m_2-m_1}{1+m_1m_2}=\\tan^{-1}\\dfrac{-\\dfrac{1}{3}-(-2)}{1+(-\\dfrac{1}{3})(-2)}"

"=\\tan^{-1}\\dfrac{5}{7}\\approx35.5\\degree"


6.



"2y = 20-3x=>y=-\\dfrac{3}{2}x+10"

"slope_1=m_1=-\\dfrac{3}{2}"

"3x + 2y = 48 =>y=-\\dfrac{3}{2}x+24"

"slope_2=m_2=-\\dfrac{3}{2}=m_1"

Two lines are parallel.

The line is perpendicular to these parallel lines


"slope_3=m_2=\\dfrac{-1}{-\\dfrac{3}{2}}=\\dfrac{2}{3}"

Then the equation of the perpendicular line is


"y=\\dfrac{2}{3}x"

"-\\dfrac{3}{2}x+10=\\dfrac{2}{3}x=>x=\\dfrac{60}{13}, y=\\dfrac{40}{13}"

"-\\dfrac{3}{2}x+24=\\dfrac{2}{3}x=>x=\\dfrac{144}{13}, y=\\dfrac{96}{13}"

"A=d^2=(\\dfrac{144}{13}-\\dfrac{96}{13})^2+(\\dfrac{60}{13}-\\dfrac{40}{13})^2"

"=\\dfrac{1904}{169} (square\\ units)"



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