Question

Find the general equation of the line parallel to 4x - 3y = 15 and passing:
a. at a distance 2 from the origin
b. twice as far from the origin
c. 2 units far from the origin
d. at a distance 5 from the given line


Find the general equation of the line parallel to x + y = 3 and passing:
a. at a distance 2 from the origin
b. 2 square root of 2 units farther from the origin
c. 2/3 as far from the origin
d. at a distance 3 square root of 2 from the given line

Find the general equation of the line:
a. parallel to the line 2x + 3y = 6 and passing at a distance 5 square root of 13 over 13 from the point (-1, 1)
b. parallel to the line x - y + 9 = 0 and passing at a distance 5 square root of 2 from the point (1, 4)
c. perpendicular to the line 3x + 4y = 7 and passing at a distance 4 from the point (1, -2)
d. perpendicular to the line 3x - 4y = 20 and passing at a distance 2 from the point (-1, 1)

Answer 1

Answer on Question #57668 - Math - Analytic Geometry

Question

1. Find the general equation of the line parallel to $4x - 3y = 15$ and passing:

a. at a distance 2 from the origin

b. twice as far from the origin

c. 2 units far from the origin

d. at a distance 5 from the given line.

Solution:

Given the line: $4x - 3y = 15$

It is equal to

$$4x - 3y - 15 = 0 \text{ or } y = \frac{4}{3}x - 5$$

Lines which are parallel to given line have the same angular coefficient $k = \frac{4}{3}$.

And their equation is $y = \frac{4}{3}x + b$, where $b$ is the intercept or $\frac{4}{3}x - y + b = 0$.

In the case of a line in the plane given by the equation $ax + by + c = 0$, where $a, b$ and $c$ are real constants with $a$ and $b$ not both zero, the distance from the line to a point $(x_0, y_0)$ is

$$d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}.$$

Let us use this formula for the next calculations.

a) At a distance 2 from the origin.

$$d = 2 \text{ from } (0; 0)$$

Substitute the relevant numbers into the formula for the distance:

$$\begin{array}{l} 2 = \frac{\left| \frac{4}{3} * 0 - 1 * 0 + b \right|}{\sqrt{\frac{16}{9} + 1}} \\ |b| = \frac{10}{3} \\ \end{array}$$ $$b = \frac{10}{3} \text{ or } b = -\frac{10}{3}$$

Substitute $b$ into $y = \frac{4}{3}x + b$.

Answer: $y = \frac{4}{3}x \pm \frac{10}{3}$.

b) Twice as far from the origin.

Given line: $4x - 3y - 15 = 0$

Point: $(0; 0)$

Find the distance from a given line to the origin with our formula:

$$d = \frac{|4 * 0 - 3 * 0 - 15|}{\sqrt{16 + 9}} = 3.$$

Twice as far from the origin: $d_{1} = 2 * d = 2 * 3 = 6$.

$$\frac {4}{3} x - y + b = 0.$$ $$d _ {1} = 6.$$

Point: (0;0).

Substitute the relevant numbers into the formula for the distance:

$$6 = \frac {\left| \frac {1 4}{3} * 0 - 1 * 0 + b \right|}{\sqrt {\frac {1 6}{9} + 1}}$$ $$| b | = 1 0$$ $$b = 1 0 \text{ or } b = - 1 0.$$

Substitute $b$ into $y = \frac{4}{3} x + b$.

Answer: $y = \frac{4}{3} x \pm 10$.

c) 2 units far from the origin.

$$d _ {1} = d + 2 = 3 + 2 = 5.$$

Point (0;0).

$$\frac {4}{3} x - y + b = 0.$$

Substitute the relevant numbers into the formula for the distance:

$$5 = \frac {\left| \frac {4}{3} * 0 - 1 * 0 + b \right|}{\sqrt {\frac {1 6}{9} + 1}}$$ $$| b | = \frac {2 5}{3}$$ $$b = \frac {2 5}{3} \text{ or } b = - \frac {2 5}{3}$$

Substitute $b$ into $y = \frac{4}{3} x + b$.

Answer: $y = \frac{4}{3} x \pm \frac{25}{3}$.

d) At a distance 5 from the given line.

At a distance $d_{1} = d + 5 = 3 + 5 = 8$ from the origin.

Point: (0;0).

$$\frac {4}{3} x - y + b = 0.$$

Substitute the relevant numbers into the formula for the distance:

$$8 = \frac {\left| \frac {4}{3} * 0 - 1 * 0 + b \right|}{\sqrt {\frac {1 6}{9} + 1}}$$ $$| b | = \frac {4 0}{3}.$$ $$b = \frac {4 0}{3} \text{ or } b = - \frac {4 0}{3}.$$

Substitute $b$ into $y = \frac{4}{3} x + b$.

Answer: $y = \frac{4}{3} x \pm \frac{40}{3}$.

Question

2. Find the general equation of the line parallel to $x + y = 3$ and passing:

a. at a distance 2 from the origin

b. 2 square root of 2 units farther from the origin

c. $\frac{2}{3}$ as far from the origin

d. at a distance 3 square root of 2 from the given line

Solution

Given the line: $x + y = 3$.

It is equal to

$$x + y - 3 = 0 \text{ or } y = -x + 3$$

Lines which are parallel to the given line have the same slope $k = -1$.

Their equation is $y = -x - b$, where $b$ is the intercept, hence obtain $x + y + b = 0$.

In the case of a line in the plane given by the equation $ax + by + c = 0$, where $a, b$ and $c$ are real constants with $a$ and $b$ not simultaneously zero, the distance from the line to a point $(x_0, y_0)$ is

$$d = \frac{\left| ax_0 + by_0 + c\right|}{\sqrt{a^2 + b^2}}.$$

Let us use this formula for the next calculations.

a) At a distance 2 from the origin.

$$d = 2 \text{ from } (0; 0).$$ $$x + y + b = 0.$$

Substitute the relevant numbers into the formula for the distance:

$$\begin{array}{l} 2 = \frac{|1 * 0 + 1 * 0 + b|}{\sqrt{1 + 1}} \\ |b| = 2\sqrt{2} \\ b = \pm 2\sqrt{2} \\ \end{array}$$

Substitute $b$ into $y = -x - b$.

Answer: $y = -x \pm 2\sqrt{2}$.

b) 2 square root of 2 units farther from the origin.

Given line: $x + y - 3 = 0$

Point: (0;0)

Find the distance from the given line to the origin with the formula:

$$d = \frac {| 1 * 0 + 1 * 0 - 3 |}{\sqrt {1 + 1}} = \frac {3}{\sqrt {2}}.$$ $$d _ {1} = d + 2 \sqrt {2} = \frac {3}{\sqrt {2}} + 2 \sqrt {2} = \frac {7}{\sqrt {2}}.$$ $$x + y + b = 0.$$ $$d _ {1} = \frac {7}{\sqrt {2}}.$$

Point: (0;0).

Substitute the relevant numbers into the formula for the distance:

$$\frac {7}{\sqrt {2}} = \frac {| 1 * 0 + 1 * 0 + b |}{\sqrt {1 + 1}}$$ $$| b | = 7$$ $$b = \pm 7$$

Substitute $b$ into $y = -x - b$.

Answer: $y = -x \pm 7$.

c) 2/3 as far from the origin.

$$d _ {1} = \frac {2}{3} d = \frac {2}{3} * \frac {3}{\sqrt {2}} = \sqrt {2}$$ $$x + y + b = 0.$$ $$d _ {1} = \sqrt {2}.$$

Point: (0;0).

Substitute the relevant numbers into the formula for the distance:

$$\sqrt {2} = \frac {| 1 * 0 + 1 * 0 + b |}{\sqrt {1 + 1}}$$ $$| b | = 2$$ $$b = \pm 2$$

Substitute $b$ into $y = -x - b$.

Answer: $y = -x \pm 2$.

d) At a distance 3 square root of 2 from the given line.

$$d _ {1} = 3 \sqrt {2} + d = 3 \sqrt {2} + \frac {3}{\sqrt {2}} = \frac {9}{\sqrt {2}}$$ $$x + y + b = 0.$$ $$d _ {1} = \frac {9}{\sqrt {2}}.$$

Point: (0;0).

Substitute the relevant numbers into the formula for the distance:

$$\frac {9}{\sqrt {2}} = \frac {| 1 * 0 + 1 * 0 + b |}{\sqrt {1 + 1}}$$ $$| b | = 9$$ $$b = \pm 9$$

Substitute $b$ into $y = -x - b$.

Answer: $y = -x \pm 9$.

Question

3. Find the general equation of the line:

a. parallel to the line $2x + 3y = 6$ and passing at a distance 5 square root of 13 over 13 from the point (-1, 1)

b. parallel to the line $x - y + 9 = 0$ and passing at a distance 5 square root of 2 from the point (1, 4)

c. perpendicular to the line $3x + 4y = 7$ and passing at a distance 4 from the point (1, -2)

d. perpendicular to the line $3x - 4y = 20$ and passing at a distance 2 from the point (-1, 1)

Solution:

a) $2x + 3y = 6$

$$y = \frac{6 - 2x}{3}$$ $$y = -\frac{2}{3}x + 2$$ $$k = -\frac{2}{3}$$ $$y = -\frac{2}{3}x + b \text{ or } -\frac{2}{3}x - y + b = 0.$$

Point: (-1, 1).

$$d = 5\sqrt{13}.$$ $$-\frac{2}{3}x - y + b = 0.$$

Substitute the relevant numbers into the formula for the distance:

$$5\sqrt{13} = \frac{\left| -1 * \left(-\frac{2}{3}\right) - 1 + b \right|}{\sqrt{\frac{4}{9} + 1}}$$ $$b = 27$$

Substitute $b$ into $y = -\frac{2}{3}x + b$

Answer: $y = -\frac{2}{3}x + 27$

b) Parallel to the line $x - y + 9 = 0$ and passing at a distance 5 square root of 2 from the point (1, 4).

$$y = x + 9$$ $$k = 1$$ $$y = x + b \text{ or } x - y + b = 0.$$

Point: (1, 4).

$$d = 5\sqrt{2}.$$

$x - y + b = 0$

Substitute the relevant numbers into the formula for the distance:

$$5 \sqrt {2} = \frac {\left| 1 * 1 - 1 * 4 + b \right|}{\sqrt {1 + 1}}$$ $$\left| - 3 + b \right| = 10$$

$b = 13$ or $b = -7$

Substitute $b$ into $y = x + b$ .

Answer: $y = x + 13$ or $y = x - 7$ .

c) Perpendicular to the line $3x + 4y = 7$ and passing at a distance 4 from the point $(1, -2)$ . If lines are perpendicular then product of their slopes is equal to $-1$ .

In the given line:

$$y = - \frac {3}{4} x + \frac {7}{4}$$ $$k = - \frac {3}{4}$$

So, the perpendicular line has a slope of $k_{1} = -\frac{1}{k} = -\frac{1}{-\frac{3}{4}} = \frac{4}{3}$ .

Equation of a perpendicular line is

$$y = \frac {4}{3} x + b$$

or

$$\frac {4}{3} x - y + b = 0.$$

Point: (1,-2).

$$d = 4.$$

Substitute the relevant numbers into the formula for the distance:

$$4 = \frac {\left| 1 * \frac {4}{3} + 2 * 1 + b \right|}{\sqrt {\frac {16}{9} + 1}}$$

$b = \frac{10}{3}$ or $b = -10$

Substitute $b$ into $y = \frac{4}{3} x + b$ .

Answer: $y = \frac{4}{3} x + \frac{10}{3}$ or $y = \frac{4}{3} x - 10$ .

d) Perpendicular to the line $3x - 4y = 20$ and passing at a distance 2 from the point $(-1, 1)$ . If lines are perpendicular then product of their slopes is equal to $-1$ .

In the given line:

$$y = \frac {3}{4} x - 5$$ $$k = \frac {3}{4}$$

So, $k_{1} = -\frac{1}{k} = -\frac{1}{\frac{3}{4}} = -\frac{4}{3}$ .

$$y = - \frac {4}{3} x + b.$$ $$- \frac {4}{3} x - y + b = 0.$$

Point: (-1,1).

$$d = 2.$$

Substitute the relevant numbers into the formula for the distance:

$$2 = \frac {\left| - \frac {4}{3} * (- 1) - 1 + b \right|}{\sqrt {\frac {16}{9} + 1}}$$ $$b = - \frac {11}{3} \text{ or } b = 3.$$

Substitute $b$ into $y = -\frac{4}{3} x + b$.

Answer: $y = -\frac{4}{3} x - \frac{11}{3}$ or $y = -\frac{4}{3} x + 3$.

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