Question

3. Find the equation of the line parallel to 5x + 6y – 10 = 0 at a distance 61 from (-3,7).

4. Find the equation of the line parallel to 3x + 4y = 20 and at a distance of 5 units from this

line.

Answer 1

3. The equation of the line parallel to $5x + 6y – 10 = 0$ is

$$5x + 6y +b = 0$$

Distance from point $(-3,7)$ to the line is

$$d=\dfrac{|5(-3)+6(7)+b|}{\sqrt{5^2+6^2}}=61$$

$$|b-27|=61\sqrt{61}$$

$$b=27+61\sqrt{61}$$

The equation of the line is

$$5x + 6y +27+61\sqrt{61} = 0$$

$$b=27-61\sqrt{61}$$

The equation of the line is

$$5x + 6y +27-61\sqrt{61} = 0$$

4. The equation of the line parallel to $3x + 4y = 20$ is

$$3x + 4y = b$$

Take the point $(4, 2)$

$$3(4) + 4(2) = 20, True$$

Distance from point $(4, 2)$ to the line $3x + 4y = b$ is

$$d=\dfrac{|3(4)+4(2)-b|}{\sqrt{3^2+4^2}}=5$$

$$|20-b|=25$$

$$b=-5$$

The equation of the line parallel to $3x + 4y = 20$ is

$$3x + 4y = -5$$

$$b=45$$

The equation of the line parallel to $3x + 4y = 20$ is

$$3x + 4y = 45$$