Question

Find the equation of the line that's passes through (1,0) and is perpendicular to y= -3x-1. Write your answer is general form.

Answer 1

Given line,

y= -3x-1

P(1,0)

Slope of the line = -3

Perpendicular slope =

$\frac{1}{3}$

Equation of line:

y-y1=m(x-x1)

$y - 0 = \frac{1}{3} (x - 1) \\ 3(y - 0) = 1(x - 1) \\ 3y = x - 1 \\ the \: equation \: is \: \\ x - 3y - 1 = 0$

Answer 2

We must must transform the standard form equation 3x+6y=5 into a slope-intercept form equation (y=mx+b) to find its slope.

3x+6y=5 (Subtract 3x on both sides.)

6y=−3x+5 (Divide both sides by 6.)

y=3/6x+5/6

y=-1/2x+5/6

The slope of our first line is equal to − 1/2  . Perpendicular lines have negative reciprocal slopes, so if the slope of one is x, the slope of the other is −1/x.

he negative reciprocal of − 1/2 is equal to 2, therefore 2 is the slope of our line.

Since the equation of line passing through the point (1,3), therefore substitute the given point in the equation y=2x+b:

3=(2×1)+b

3=2+b

b=3−2=1

Substitute this value for b in the equation y=2x+b:

y=2x+1

Hence, the equation of the line is y=2x+1.