Question

In Exercises 13 - 16 , write an equation of the line passing through potat that parallel the given Graph the equations of the lines to check that they are parallel

13. P(0, - 1), y = - 2x + 3

14. P(3, 8), y = 1/5 (x + 4)

15. P(- 2, 6), x = - 5

16. P(4, 0) , - x + 2y = 12

In Exercises 17 - 20 , write an equation of the line passing through point P that is perpendicular to the given line. Graph the equations of the lines to check that they are perpendlezlar.

17. P(0, 0), y = - 9x - 1

18. P(4, - 6), y = - 3

19. P(2, 3), y - 4 = - 2(x + 3)

20. P(- 8, 0), 3x - 5y = 6

Answer 1

Linear equations can be parallel, perpendicular or have no relationship at all.

Question 13 to 16: Parallel graphs

13. P(0,-1), y = -2x + 3

Parallel lines have the same slope.

So, the slope (m) is: -2

The equation is then calculated as:

$\mathbf{y = m(x - x_1) + y_1}$

This gives

$\mathbf{y = -2(x - 0) - 1}$

$\mathbf{y = -2x - 1}$

Hence, the equation is: $\mathbf{y = -2x - 1}$

14. P(3,8), y = 1/5(x + 4)

The slope (m) is: 1/5

The equation is then calculated as:

$\mathbf{y = m(x - x_1) + y_1}$

This gives

$\mathbf{y = \frac 15(x - 3) + 8}$

$\mathbf{y = \frac 15x - \frac 35 + 8}$

$\mathbf{y = \frac 15x + \frac{-3 + 40}5}$

$\mathbf{y = \frac 15x + \frac{37}5}$

Hence, the equation is: $\mathbf{y = \frac 15x + \frac{37}5}$

15. P(-2,6), x = -5

The graph of x = -5 is a vertical line that passes through point x = -5.

Vertical lines have undefined slopes

Hence, the equation is: x = -2

16. P(4,0), -x + 2y = 12

The slope (m) is: 6

The equation is then calculated as:

$\mathbf{y = m(x - x_1) + y_1}$

This gives

$\mathbf{y = 6(x - 4) + 0}$

$\mathbf{y = 6x - 24}$

Hence, the equation is: $\mathbf{y = 6x - 24}$

Question 17 to 29: Perpendicular graphs

17. P(0,0), y = -9x - 1

The relationship between the perpendicular graphs is: $\mathbf{m_2 = -\frac{1}{m_1}}$

So, the slope (m) is: 1/9

The equation is then calculated as:

$\mathbf{y = m(x - x_1) + y_1}$

This gives

$\mathbf{y = \frac 19(x - 0) + 0}$

$\mathbf{y = \frac 19x}$

Hence, the equation is: $\mathbf{y = \frac 19x}$

18. P(4, -6), y = -3

The graph of y = -3 is a horizontal line that passes through y = -3

The slope of y = -3 is 0

So, the slope of the required graph is undefined

Hence, the equation is: x = 4

19. P(2,3), y - 4 = -2(x + 3)

The slope (m) is: 1/2

The equation is then calculated as:

$\mathbf{y = m(x - x_1) + y_1}$

This gives

$\mathbf{y = \frac 12(x - 2) + 3}$

$\mathbf{y = \frac 12x - 1 + 3}$

$\mathbf{y = \frac 12x + 2}$

Hence, the equation is: $\mathbf{y = \frac 12x + 2}$

16. P(-8,0), 3x - 5y = 6

The slope (m) is: -5/3

The equation is then calculated as:

$\mathbf{y = m(x - x_1) + y_1}$

This gives

$\mathbf{y = -\frac 53(x + 8) + 0}$

$\mathbf{y = -\frac 53(x + 8)}$

Hence, the equation is: $\mathbf{y = -\frac 53(x + 8)}$

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