The equation of line WX is 2x + y = −5. What is the equation of a line perpendicular to line WX in slope-intercept form that contains point (−1, −2)?

Respuesta :

Answer: [tex]y=\dfrac12x-\dfrac{3}{4}[/tex]

Step-by-step explanation:

Given, The equation of line WX is 2x + y = −5.

It can be written as [tex]y=-2x-5[/tex] comparing it with slope-intercept form y=mx+c, where m is slope and c is y-intercept, we have

slope of WX = -2

Product of slopes of two perpendicular lines is -1.

So, (slope of WX) × (slope of perpendicular to WX)=-1

[tex]-2\times\text{slope of WX}=-1\\\\\Rightarrow\ \text{slope of WX}=\dfrac{1}{2}[/tex]

Equation of a line passes through (a,b) and has slope m:

[tex]y-b=m(x-a)[/tex]

Equation of a line perpendicular to WX contains point (−1, −2) and has slope [tex]=\dfrac12[/tex]

[tex]y-(-2)=\dfrac{1}{2}(x-(-1))\\\\\Rightarrow\ y+2=\dfrac12(x+1)\\\\\Rightarrow\ y+2=\dfrac12x+\dfrac12\\\\\Rightarrow\ y=\dfrac12x+\dfrac12-2\\\\\Rightarrow\ y=\dfrac12x-\dfrac{3}{4}[/tex]

Equation of a line perpendicular to line WX in slope-intercept form that contains point (−1, −2) [tex]:y=\dfrac12x-\dfrac{3}{4}[/tex]