Question

Write an equation in slope intercept form of the line passing through the given point in perpendicular to the given line.16. (-1,2); y= 3x + 118. (3, 0); X - y = 417. （4,2);x+y=119. （-7,-3);2x+4y=8

Answer 1

Given that a line passes through the point (-1, 2) and is perpendicular to line y = 3x + 1

The general form of the equation of a line is

`\begin{gathered} y=mx+b \\ \text{Where m is the slope and b is the y-intercept} \end{gathered}`

Let m₁ be the slope of the given line

The slope, m₁, of the given line y = 3x + 1 is

`m_1=3`

Since the line passing through the point is perpendicular to the given line, the formula to find the slope of perpendicular lines is

`m_2=-(1)/(m_1)`

Let m₂ be the slope of the line perpendicular to the given line.

The slope, m₂, of the line perpendicular to the given line is

`\begin{gathered} m_2=-(1)/(m_1) \\ \text{Where } \\ m_1=3 \\ m_2=-(1)/(3)_{} \end{gathered}`

The slope, m₂, of the line perpendicular to the given line is -1/3

To find the equation of a straight line, the formula is

`(y-y_1)/(x-x_1)=m_{}`

Where

`\begin{gathered} m=m_2=-(1)/(3) \\ (x_1,y_1)\Rightarrow(-1,2) \end{gathered}`

Substitute the values into the formula to find the equation of a straight line

`\begin{gathered} (y-y_1)/(x-x_1)=m_2 \\ (y-2)/(x-(-1))=-(1)/(3) \\ (y-2)/(x+1)=-(1)/(3) \end{gathered}`

Crossmultiply

Hence, the slope-intercept form of the line is

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