Question

1a) what would be the gradient of a line parallel to the straight line 3x-y+4=0

b) what would be the gradient of a line perpendicular to the straight line
5x-2y-1=0

2. State whether the following pairs of lines whose equations are given are parallel, perpendicular or neither.

a. 2x-y+4=0 and 6x-3y+7=0
b. 7x+3y-8=0 and 3x-7y+1=0
c. x+3y-2=0 and 3x-y+4=0

3. Find the equation of the straight lines

a) Passing through the point (3, -2) and parallel to the line 4x-y+6=0

b) Passing through the origin and parallel to the line 5x+3y-7=0

c) Passing through the point (-2,5) and perpendicular to the line 3x-2y+8=0

Answer 1

Answer on Question #78301 – Math – Linear Algebra

Question

1a) what would be the gradient of a line parallel to the straight line

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

Solution

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

The equation of the straight line in slope-intercept form

$$y = 3x + 4$$

The gradient of a line parallel to the given straight line is

$$grad = m = 3$$

Answer: 3.

Question

b) what would be the gradient of a line perpendicular to the straight line

$$5x - 2y - 1 = 0$$

Solution

$$5x - 2y - 1 = 0$$

The equation of the straight line in slope-intercept form

$$y = \frac{5}{2}x - \frac{1}{2}$$

If two lines are perpendicular

$$grad_1 \cdot grad_2 = -1$$

The gradient of a line perpendicular to the given straight line is

$$grad_2 = -\frac{1}{grad_1} = -\frac{1}{\frac{5}{2}} = -\frac{2}{5}$$

Answer: $-\frac{2}{5}$.

2. State whether the following pairs of lines whose equations are given are parallel, perpendicular or neither.

Question

a. $2x - y + 4 = 0$ and $6x - 3y + 7 = 0$

Solution

$2x - y + 4 = 0$ and $6x - 3y + 7 = 0$

The equations of the straight line in slope-intercept form

$$y = 2x + 4 \quad \text{and} \quad y = 2x + \frac{7}{3}$$ $$grad_1 = 2 = grad_2$$

Therefore, two lines are parallel.

Answer: two lines are parallel.

Question

b. $7x + 3y - 8 = 0$ and $3x - 7y + 1 = 0$

Solution

$7x + 3y - 8 = 0$ and $3x - 7y + 1 = 0$

The equations of the straight line in slope-intercept form

$$\begin{array}{l} y = - \frac {7}{3} x + \frac {8}{3} \quad \text{and} \quad y = \frac {3}{7} x + \frac {1}{7} \\ \operatorname{grad}_1 = - \frac {7}{3}, \operatorname{grad}_2 = \frac {3}{7} \\ \operatorname{grad}_1 \cdot \operatorname{grad}_2 = - \frac {7}{3} \cdot \frac {3}{7} = - 1 \end{array}$$

Therefore, two lines are perpendicular.

Answer: two lines are perpendicular.

Question

c. $x + 3y - 2 = 0$ and $3x - y + 4 = 0$

Solution

$x + 3y - 2 = 0$ and $3x - y + 4 = 0$

The equations of the straight line in slope-intercept form

$$\begin{array}{l} y = - \frac {1}{3} x + \frac {2}{3} \quad \text{and} \quad y = 3 x + 4 \\ \operatorname{grad}_1 = - \frac {1}{3}, \operatorname{grad}_2 = 3 \\ \operatorname{grad}_1 \cdot \operatorname{grad}_2 = - \frac {1}{3} \cdot 3 = - 1 \end{array}$$

Therefore, two lines are perpendicular.

Answer: two lines are perpendicular.

3. Find the equation of the straight lines

Question

a) Passing through the point $(3, -2)$ and parallel to the line $4x - y + 6 = 0$

Solution

$4x - y + 6 = 0$

The equation of the straight line in slope-intercept form

$$y = 4 x + 6$$

Two lines are parallel, then

$$grad_1 = grad_2 = 4$$

The equation of new straight line in slope-intercept form

$$y = grad_2 \cdot x + b_2$$ $$y = 4x + b_2$$

This line passes through the point $(3, -2)$

$$-2 = 4(3) + b_2$$ $$b_2 = -14$$

The equation of new straight line in slope-intercept form

$$y = 4x - 14$$

The equation of new straight line in general form

$$4x - y - 14 = 0$$

Answer: $4x - y - 14 = 0$.

Question

b) Passing through the origin and parallel to the line $5x + 3y - 7 = 0$

Solution

$$5x + 3y - 7 = 0$$

The equation of the straight line in slope-intercept form

$$y = -\frac{5}{3}x + \frac{7}{3}$$

Two lines are parallel, then

$$grad_1 = grad_2 = -\frac{5}{3}$$

The equation of a new straight line in slope-intercept form

$$y = grad_2 \cdot x + b_2$$ $$y = -\frac{5}{3}x + b_2$$

This line passes through the origin

$$0 = -\frac{5}{3}(0) + b_2$$ $$b_2 = 0$$

The equation of a new straight line in slope-intercept form

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

The equation of a new straight line in general form

$$5x + 3y = 0$$

Answer: $5x + 3y = 0$.

Question

c) Passing through the point $(-2,5)$ and perpendicular to the line

$$3x - 2y + 8 = 0$$

**Solution**

$$3x - 2y + 8 = 0$$

The equation of the straight line in slope-intercept form

$$y = \frac{3}{2}x + 4$$

Two lines are perpendicular, then

$$grad_1 \cdot grad_2 = -1$$ $$\frac{3}{2} \cdot grad_2 = -1$$ $$grad_2 = -\frac{2}{3}$$

The equation of new straight line in slope-intercept form

$$y = grad_2 \cdot x + b_2$$ $$y = -\frac{2}{3}x + b_2$$

This line passes through the point $(-2,5)$

$$5 = -\frac{2}{3}(-2) + b_2$$ $$b_2 = \frac{11}{3}$$

The equation of new straight line in slope-intercept form

$$y = -\frac{2}{3}x + \frac{11}{3}$$

The equation of new straight line in general form

$$2x + 3y - 11 = 0$$

**Answer:** $2x + 3y - 11 = 0$.

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