Question

Determine the equations of the following lines:

1. Parallel to x -3y = 9 and passing through the point (2;6)

2. Perpendicular to y + 1/4x -5 = 0 and passing though that point (-3;5)

Answer 1

1) x - 3y = -16

2) 4x - y = -17

Explanation:

Question 1

To determine the equation of a line that is parallel to x - 3y = 9 and passes through the point (2, 6), we first need to find the slope of x - 3y = 9.

To do this, rearrange the equation so that it is in slope-intercept form.

Slope intercept form is y = mx + b, where m is the slope and b is the y-intercept.

`\begin{aligned}x-3y&=9\\x-3y+3y-9&=9+3y-9\\x-9&=3y\\3y&=x-9\\y&=(1)/(3)x-3\end{aligned}`

Therefore, the slope of the given line is 1/3.

Parallel lines have the same slope.

Therefore, to find the equation of the parallel line that passes through point (2, 6), substitute m = 1/3 and the point (2, 6) into the point-slope formula:

`\begin{aligned}y-y_1&=m(x-x_1)\\\\\implies y-6&=(1)/(3)(x-2)\end{aligned}`

Rearrange to standard form Ax + By = C (where A is positive):

`\begin{aligned}y-6&=(1)/(3)(x-2)\\3y-18&=x-2\\-18&=x-3y-2\\x-3y&=-16\end{aligned}`

Therefore, the equation of the line in standard form that is parallel to x - 3y = 9 and passes through the point (2, 6) is:

`\boxed{x-3y=-16}`

`\hrulefill`

Question 2

To determine the equation of a line that is perpendicular to y + 1/4x - 5 = 0 and passes through the point (-3, 5), we first need to find the slope of y + 1/4x - 5 = 0.

To do this, rearrange the equation so that it is in slope-intercept form.

Slope intercept form is y = mx + b, where m is the slope and b is the y-intercept.

`\begin{aligned}y + (1)/(4)x - 5 &= 0\\y&=-(1)/(4)x+5 \end{aligned}`

Therefore, the slope of the given line is -1/4.

The slopes of perpendicular lines are negative reciprocals.

Therefore, the slope of the perpendicular line is 4.

Therefore, to find the equation of the perpendicular line that passes through point (-3, 5), substitute m = 4 and the point (-3, 5) into the point-slope formula:

`\begin{aligned}y-y_1&=m(x-x_1)\\\\\implies y-5&=4(x-(-3))\end{aligned}`

Rearrange to standard form Ax + By = C (where A is positive):

`\begin{aligned}y-5&=4(x-(-3))\\y-5&=4(x+3)\\y-5&=4x+12\\-5-12&=4x-y\\4x-y&=-17\end{aligned}`

Therefore, the equation of the line in standard form that is perpendicular to y + 1/4x - 5 = 0 and passes though point (-3, 5) is:

`\boxed{4x-y=-17}`