Question

Write an equation in slope-intercept form of a line passingthrough the given point and perpendicular to the given line.Write an equation in slope-intercept forrthrough the given point and perpendicular ~ (-5,1); y= -X

Answer 1

Solution

The formula to find the equation of a straight line is

$m=\frac{y-y_1}{x-x_1}$

Given that the line passes through the point (-5,1) which is parallel to the given line

$y=-x$

Since the lines are perpendicular, then to find the slope of the other line, the formula is

$m_2=-\frac{1}{m_1}$

Let m₁ be the slope of the given line and m₂ be the slope of the line perpendicular to the given line

The general form of an equation of a straight line is

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

The slope, m₁, of the given line is -1, i.e m₁ = -1

The slope, m₂, of the line perpedicular to the given line will be

$\begin{gathered} m_1=-1 \\ m_2=-\frac{1}{m_1} \\ m_2=-\frac{1}{-1} \\ m_2=1 \end{gathered}$

The slope, m₂, of the line passing through the points (-5, 1), m₂ = 1

Where

$\begin{gathered} (x_1,y_1)\Rightarrow(-5,1) \\ m_2=1 \end{gathered}$

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

$\begin{gathered} m=\frac{y-y_1}{x-x_1} \\ 1=\frac{y-1}{x-(-5)} \\ 1=\frac{y-1}{x+5} \\ \text{Crossmultiply} \\ x+5=y-1 \\ y=x+5+1 \\ y=x+6 \end{gathered}$

Hence, the equation of the line in slope-intercept form is

$y=x+6$