Question

Find the general equation of the line passing through the given point and parallel to the indicated equation of the line:
a. (4, 1); 3X + 4Y - 10 = 0
b. (-1/2, -4); 7x - 8y - 5 = 0

Find the general equation of the line passing through the given point and perpendicular to the indicated equation of the line
a. (, 4); 4x + 4y - 11 = 0
b. (-4, -1/3); 7x - 8y -5 = 0

Find the angle of inclination of a line whose equation is
a. 6x - 5y + 30 = 0
b. 3x - 5y + 6 = 0
c. 12x - 9y = 32

Answer 1

Answer on Question #57670 - Math - Analytic Geometry

1. Find the general equation of the line passing through the given point and parallel to the indicated equation of the line:

a. $(4, 1); 3x + 4y - 10 = 0$

b. $(-1/2, -4); 7x - 8y - 5 = 0$

Solution:

a. $(4, 1); 3x + 4y - 10 = 0$

The general equation is $Ax + By + C = 0$. If lines are parallel we can use $A = 3$, $B = 4$.

$$3x + 4y + C = 0$$

If line passes through the given point that the coordinates of point correlate with general equation:

$$3 \cdot 4 + 4 \cdot 1 + C = 0 \rightarrow C = -16.$$

The general equation is $3x + 4y - 16 = 0$.

b. $(-1/2, -4); 7x - 8y - 5 = 0$

We solve in the same way.

$$7x - 8y + C = 0 \rightarrow 7 \cdot \left(-\frac{1}{2}\right) - 8 \cdot (-4) + C = 0 \rightarrow C = -28.5$$

The general equation is $7x - 8y - 28.5 = 0$

2. Find the general equation of the line passing through the given point and perpendicular to the indicated equation of the line

a. $(, 4); 4x + 4y - 11 = 0$

b. $(-4, -1/3); 7x - 8y - 5 = 0$

Solution:

The Slope-Intercept Form of the equation of a straight line is $y = mx + b$.

If line1 and line2 are perpendicular then $m_1 \cdot m_2 = -1$

a. $(, 4); 4x + 4y - 11 = 0$

$$\begin{array}{l} \text{Line1: } 4x + 4y - 11 = 0 \rightarrow 4y = 11 - 4x \rightarrow y = -x + \frac{11}{4} \\ m_1 = -1 \rightarrow m_2 = 1 \\ \end{array}$$

Line 2: $y = x + b$.

We have a point of line2 $(0,4)$: $4 = 0 + b$, $b = 4$

$$y = x + 4$$

The general equation is $x - y + 4 = 0$

b. $(-4, -1/3); 7x - 8y - 5 = 0$

$$\begin{array}{l} \text{Line1: } 7x - 8y - 5 = 0 \rightarrow 8y = 7x - 5 \rightarrow y = \frac{7}{8}x - \frac{5}{8} \\ m_1 = \frac{7}{8} \rightarrow m_2 = -\frac{8}{7} \\ \end{array}$$

Line 2: $y = -\frac{8}{7} + b$.

We have a point of line2 $(-4, -1/3)$: $-\frac{1}{3} = -\frac{8}{7} \cdot (-4) + b$, $b = \frac{89}{21}$

$$y = -\frac{8}{7}x + \frac{89}{21} \rightarrow y + \frac{8}{7}x - \frac{89}{21} = 0 \rightarrow 24x + 21y - 89 = 0$$

The general equation is $24x + 21y - 89 = 0$

Find the angle of inclination of a line whose equation is

a. $6x - 5y + 30 = 0$

b. $3x - 5y + 6 = 0$

c. $12x - 9y = 32$

**Solution:**

The Slope-Intercept Form of the equation of a straight line is $y = mx + b$.

$$m = \tan \theta$$

a. $6x - 5y + 30 = 0$

$$\begin{array}{l} 6x - 5y + 30 = 0 \rightarrow y = \frac{6}{5}x + 6 \rightarrow y = 1.2x + 6 \\ \tan \theta = 1.2 \\ \theta \approx 50{}^\circ \\ \end{array}$$

b. $3x - 5y + 6 = 0$

$$\begin{array}{l} 3x - 5y + 6 = 0 \rightarrow y = \frac{3}{5}x + \frac{6}{5} \rightarrow y = 0.6x + \frac{6}{5} \\ \tan \theta = 0.6 \\ \theta \approx 31{}^\circ \\ \end{array}$$

c. $12x - 9y = 32$

$$\begin{array}{l} 12x - 9y = 32 \rightarrow y = \frac{12}{9}x - \frac{32}{9} \rightarrow y = \frac{4}{3}x - \frac{32}{9} \\ \tan \theta \approx 1.3333 \\ \theta \approx 53.1{}^\circ \\ \end{array}$$

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