Question

Find the angle formed from line 1 to line 2:
a. line 1: 4x - 5y + = 0; line 2: 6x -4y - 12 = 0
b. line 1: 2x + 7y = 0; line 2: 3x - 5y - 15 = 0

Find the equation of the line:
1. having an x-intercept 4 and slope 5
2. through (5, -8) and with intercepts equal
3. through (-6, 3) and with intercepts numerically equal but opposite in sign
4. through (-5, 3) and with x-intercept twice the y-intercept
5. through (3, 2) and having a slope equal to two-thirds of its y-intercept
6. through (-4, -2) and with sum of intercepts 3
7. through (4, -2) and with product of intercepts -18.
8. through (4, -2) and forming with the axes a triangle of area 9 square units

Answer 1

Answer the question #57669 – Mathematics – Analytic Geometry

Find the angle formed from line 1 to line 2:

a. line 1: $4x - 5y \neq 0$; line 2: $6x - 4y - 12 = 0$ (may be a bug!)

b. line 1: $2x + 7y = 0$; line 2: $3x - 5y - 15 = 0$

Solution.

a. Let $\alpha$ be an angle formed from line 1 to line 2, then (by corresponding formula)

$$\cos \alpha = \frac {4 \cdot 6 + (- 5) \cdot (- 4)}{\sqrt {4 ^ {2} + (- 5) ^ {2}} \cdot \sqrt {6 ^ {2} + (- 4) ^ {2}}} = \frac {2 4 + 2 0}{\sqrt {1 6 + 2 5} \cdot \sqrt {3 6 + 1 6}} = \frac {4 4}{\sqrt {4 1 \cdot 5 2}} = \frac {2 2}{\sqrt {5 3 3}}$$

and $\alpha = \arccos \left(\frac{22}{\sqrt{533}}\right) \approx 17.65$ degrees.

b. Let $\alpha$ be an angle formed from line 1 to line 2, then (by corresponding formula)

$$\cos \alpha = \frac {2 \cdot 3 + 7 \cdot (- 5)}{\sqrt {2 ^ {2} + 7 ^ {2}} \cdot \sqrt {3 ^ {2} + (- 5) ^ {2}}} = \frac {6 - 3 5}{\sqrt {4 + 4 9} \cdot \sqrt {9 + 2 5}} = \frac {- 2 9}{\sqrt {5 3 \cdot 3 4}} = - \frac {2 9}{\sqrt {1 8 0 2}}$$

and $\alpha = \arccos \left(\frac{29}{\sqrt{1802}}\right) \approx 47$ degrees.

Answer: found.

Find the equation of the line:

1. having an $x$-intercept 4 and slope 5

2. through (5, -8) and with intercepts equal

3. through $(-6, 3)$ and with intercepts numerically equal but opposite in sign

4. through $(-5, 3)$ and with $x$-intercept twice the $y$-intercept

5. through $(3,2)$ and having a slope equal to two-thirds of its $y$-intercept

6. through $(-4, -2)$ and with sum of intercepts 3

7. through $(4, -2)$ and with product of intercepts -18.

8. through $(4, -2)$ and forming with the axes a triangle of area 9 square units.

Solution.

1. We have $y = 5x + b$, and $0 = 5 \cdot 4 + b$. It follows that $b = -20$ and the equation of this line: $5x - y - 20 = 0$.

2. We have $\frac{x}{a} + \frac{y}{a} = 1$, and $\frac{5}{a} - \frac{8}{a} = 1$. It follows that $a = -3$ and the equation of this line is $-\frac{x}{3} - \frac{y}{3} = 1$ or $x + y + 3 = 0$.

3. We have $\frac{x}{a} - \frac{y}{a} = 1$, and $-\frac{6}{a} - \frac{3}{a} = 1$. It follows that $a = -9$ and the equation of this line is $-\frac{x}{9} + \frac{y}{9} = 1$ or $x - y + 9 = 0$.

4. We have $\frac{x}{2a} + \frac{y}{a} = 1$, and $-\frac{5}{2a} + \frac{3}{a} = 1$. It follows that $a = \frac{1}{2}$ and the equation of this line is $x + 2y - 1 = 0$.

5. We have $y = \frac{2ax}{3} + a$, and $2 = \left(\frac{4}{3} + 1\right)a$. It follows that $a = \frac{6}{7}$ and the equation of this line is $y = \frac{4}{7}x + \frac{6}{7}$ or $4x - 7y + 6 = 0$.

6. We have $- + \frac{y}{3 - a} = 1$ , and $- \frac{4}{3 - a} - \frac{2}{3 - a} = 1$ , $a^2 - a - 12 = 0$ . It follows that: 1) $a = 4$ and the equation of this line is $\frac{4}{4} - y = 1$ or $\underline{x - 4y - 4 = 0}$ , or 2) $a = -3$ and the equation of this line is $-\frac{3}{4} + \frac{6}{4} = 1$ or $2x - y + 6 = 0$ .

7. We have $- - \frac{18y}{40} = 1$ , and $\frac{4}{4} + \frac{36}{40} = 1$ . It follows that $a = 40$ and the equation of this line is $\frac{18y}{40} = 1$ or $\underline{x - 18y - 40 = 0}$ .

8. So as the line is forming with the axes a triangle of area 9 square units, it means that line has the absolute value of product of intercepts be equal to 18. We have the next two cases:

1) $- + \frac{18y}{32} = 1$ , and $\frac{4}{32} - \frac{36}{32} = 1$ . It follows that $a = -32$ and the equation of this line is $-\frac{18y}{32} - \frac{18y}{32} = 1$ or $\underline{x + 18y + 32 = 0}$ .

2) $- - \frac{18y}{40} = 1$ , and $\frac{4}{4} + \frac{36}{40} = 1$ . It follows that $a = 40$ and the equation of this line is $\frac{18y}{40} = 1$ or $\underline{x - 18y - 40 = 0}$ .

Answer: all was found.

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