Question

Write an equation of the line that passes through the point (1,5) and is (a) parallel to the line y = 3x-5 and (b) perpendicular

to the line y = 3x - 5.

Answer 1

`\textsf{An equation of the parallel line is $y = \boxed{3x+2}$}`

`\textsf{An equation of the perpendicular line is $y = \boxed{-(1)/(3)x+(16)/(3)}$}`

Explanation:

Point-slope form of a linear equation:

`\boxed{y-y_1=m(x-x_1)}`

Where:

• m is the slope.
• (x₁, y₁) is a point on the line.

Given:

• Equation: y = 3x - 5
• Point: (1, 5)

Part (a)

If two lines are parallel, they have the same slope.

Therefore, the slope of the line that is parallel to the given equation is 3.

To find the equation of the line, substitute the found slope and the given point into the point-slope form of a linear equation:

`\implies y-y_1=m(x-x_1)`

`\implies y-5=3(x-1)`

`\implies y-5=3x-3`

`\implies y-5+5=3x-3+5`

`\implies y=3x+2`

Part (b)

If two lines are perpendicular, their slopes are negative reciprocals.

Therefore, the slope of the line that is perpendicular to the given equation is -¹/₃.

To find the equation of the line, substitute the found slope and the given point into the point-slope form of a linear equation:

`\implies y-y_1=m(x-x_1)`

`\implies y-5=-(1)/(3)(x-1)`

`\implies y-5=-(1)/(3)x+(1)/(3)`

`\implies y-5+5=-(1)/(3)x+(1)/(3)+5`

`\implies y=-(1)/(3)x+(16)/(3)`