Question

Find the equation of the following lines:

parallel to the line joining (1;2) and (-2;-2) and passing through (4;1)

passing through the point (2; -3) and perpendicular to the line joining (2;-3) to (-1;-1)

Answer 1

`\textsf{1)}\quad y = (4)/(3)x-(13)/(3)`

`\textsf{2)} \quad y = (3)/(2)x-6`

Explanation:

To find the equation of a line parallel to the line joining (1, 2) and (-2, -2) and passing through (4, 1), we first need to find the slope of the line joining (1, 2) and (-2, -2).

`\textsf{slope}\:(m)=(y_2-y_1)/(x_2-x_1)=(-2-2)/(-2-1)=(-4)/(-3)=(4)/(3)`

Parallel lines have the same slope, so the slope of the parallel line is m = 4/3.

Substitute the found slope and point (4, 1) into the point-slope formula:

`\begin{aligned} y - y_1 &= m(x - x_1)\\\\y - 1 &= (4)/(3)(x - 4)\\\\y - 1 &= (4)/(3)x-(16)/(3)\\\\y &= (4)/(3)x-(13)/(3)\end{aligned}`

Therefore, the equation of the line parallel to the line joining (1, 2) and (-2, -2) and passing through (4, 1) is:

`\boxed{y = (4)/(3)x-(13)/(3)}`

`\hrulefill`

To find the equation of a line perpendicular to the line joining (2, -3) and (-1, -1) and passing through (2, -3), we first need to find the slope of the line joining (2, -3) and (-1, -1).

`\textsf{slope}\:(m)=(y_2-y_1)/(x_2-x_1)=(-1-(-3))/(-1-2)=(2)/(-3)=-(2)/(3)`

The slopes of perpendicular lines are negative reciprocals, so the slope of the parallel line is m = 3/2.

Substitute the found slope and point (2, -3) into the point-slope formula:

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

Therefore, the equation of the line perpendicular to the line joining (2, -3) and (-1, -1) and passing through (2, -3):

`\boxed{y = (3)/(2)x-6}`