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3. Find the standard form of a line through (1/5,2/3) that is perpendicular to

the line 4x-8y=0

User Mechenbier
by
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1 Answer

13 votes
13 votes

Answer:

15y + 30x = 16

Explanation:

The first line is 4x - 8y = 0

Convert this to slope-intercept form y = mx + b where m = slope

Multiplying by -1 ==> -4x + 8y = 0

Add -4x to both sides ==> 8y = 4x

Divide both sides by 8

==> y = (4/8)x = (1/2)x (L1)
Slope = 1/2 and y-intercept = 0

Let L2 represent the equation of the line perpendicular to L1

The slope of a line, L2, perpendicular to line L1 will have as its slope the negative of the reciprocal of slope of L1
~ Reciprocal of L1 slope = 1 ÷ 1/2 = 2
~ Negative of reciprocal = -2

Slope of L2 = -2

Equation of L2 is y = -2x + b

To find b, substitute point (x = 1/5, y = 2/3) into L2 equation and solve for b

We get 2/3 = -2(1/5) + b

2/3 = - 2/5 + b

Add 2/5 to both sides
2/3 + 2/5 = -2/5 + 2/5 + b
2/3 + 2/5 = b

or

b = 2/3 + 2/5

Multiply both sides by 15 (LCM of 3 and 5)
15b = 15(2/3) + 15(2/5) = 10 + 6 = 16

15b = 16

b = 16/15

So equation of L2, the line perpendicular to L1 is

y = -2x + 16/15

This is in slope intercept form. We have to convert it to standard form which is Ax + by = C

y = -2x + 16/15

Add 2x to both sides:

y + 2x = 2x - 2x + 16/15 = 16/15

Multiply throughout by 15

==> 15.y + 15.2x =16/15(15)

==> 15y + 30x = 16

User Guy Blanc
by
3.1k points