Question

Draw lines/graph for each item;

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.

Answer 1

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

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

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

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

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

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