Question

Write an equation that goes through (8,1) and is perpendicular to 2y + 4x =12

Answer 1

2y -x + 6 = 0

Explanation:

Here a equation and a point is given to us and we are interested in finding a equation which is perpendicular to the given equation and passes through the given point .

The given equation is ,

`\sf \longrightarrow 2y + 4x = 12 \\`

`\sf \longrightarrow 4x + 2y = 12`

Firstly convert this into slope intercept form , to find out the slope of the line.

`\sf \longrightarrow 2y = -4x +12\\`

`\sf \longrightarrow y =(-4x+12)/(2)\\`

`\sf \longrightarrow y =(-4x)/(2)+(12)/(2)\\`

`\sf \longrightarrow \red{ y = -2x + 6 }`

Now on comparing it to slope intercept form which is y = mx + c , we have ,

• m = -2

And as we know that the product of slopes of two perpendicular lines is -1 . So the slope of the perpendicular line will be negative reciprocal of the slope of the given line. Therefore ,

`\sf \longrightarrow m_(\perp)=(-1)/(-2)=\red{(1)/(2)}`

Now we may use the point slope form of the line to find out the equation of the line using the given point . The point slope form is,

`\sf \longrightarrow y - y_1 = m(x - x_1)`

Now on substituting the respective values we have,

`\sf \longrightarrow y - 1 = (1)/(2)(x-8)\\`

`\sf \longrightarrow 2(y-1 )= x -8 \\`

`\sf \longrightarrow 2y -2=x-8\\`

`\sf \longrightarrow \underline{\boxed{\bf 2y - x + 6=0}}`