Question

Find the equation of the line which is twice as far from the line 3x+4y-6=0 as from 4x+3y-5=0

Answer 1

Answer on Question #58512 – Math – Analytic Geometry

Question

Find the equation of the line which is twice as far from the line $4x + 3y - 6 = 0$ as from $4x + 3y - 5 = 0$

Solution

Assume line 1: $4x + 3y - 6 = 0$, line 2: $4x + 3y - 5 = 0$.

All three lines should be parallel, so equation of the line have the following form:

$$4x + 3y - d = 0.$$

Distance to line 1:

$$d_1 = \frac{|6 - d|}{\sqrt{4^2 + 3^2}} = \frac{|6 - d|}{5}$$

Distance to line 2:

$$d_2 = \frac{|5 - d|}{\sqrt{4^2 + 3^2}} = \frac{|5 - d|}{5}$$

We should find the line that satisfies the equality

$$d_1 = 2d_2$$

Substitute for $d_1$ and $d_2$:

$$\frac{|6 - d|}{5} = 2\frac{|5 - d|}{5}$$ $$|6 - d| = 2|5 - d|$$

Solving this equation obtain solutions:

$$\left[ \begin{array}{l} d = \frac{16}{3} \\ d = 4 \end{array} \right.$$

**Answer**: Two possible solutions are $4x + 3y - 4 = 0$ or $4x + 3y - \frac{16}{3} = 0$.

Question

Find the equation of the line which is twice as far from the line $3x + 4y - 6 = 0$ as from $3x + 4y - 5 = 0$

Solution

Assume line 1: $3x + 4y - 6 = 0$, line 2: $3x + 4y - 5 = 0$.

All three lines should be parallel, so equation of the line have the following form:

$$3x + 4y - d = 0.$$

Distance to line 1:

$$d_1 = \frac{|6 - d|}{\sqrt{4^2 + 3^2}} = \frac{|6 - d|}{5}$$

Distance to line 2:

$$d _ {2} = \frac {| 5 - d |}{\sqrt {4 ^ {2} + 3 ^ {2}}} = \frac {| 5 - d |}{5}$$

We should find the line that satisfies the equality

$$d _ {1} = 2 d _ {2}$$

Substitute for $d_{1}$ and $d_{2}$ :

$$\frac {| 6 - d |}{5} = 2 \frac {| 5 - d |}{5}$$ $$| 6 - d | = 2 | 5 - d |$$

Solving this equation obtain solutions:

$$\left[ \begin{array}{l} d = \frac {1 6}{3} \\ d = 4 \end{array} \right.$$

Answer: Two possible solutions are $3x + 4y - 4 = 0$ or $3x + 4y - \frac{16}{3} = 0$ .

www.AssignmentExpert.com