What is the equation of the line perpendicular to y=−58x that passes through (−6,3)?

2 Answers
Tony B
Jan 16, 2016

y=85x+12610

Explanation:

Consider the standard equation form of a strait line graph:

y=mx+c where m is the gradient.

A straight line that is perpendicular to this will have the gradient: −1m

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Find the generic equation of the line perpendicular to the original

Given equation: y1=−58x...............................(1)

The equation perpendicular to this will be

××××y2=+85x+c......................................(2)

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To find the value of the constant

We know it passes through the point (x,y)→(−6,3)

Substitute this point into equation (2) giving:

y2=3=85(−6)+c

y2=3=−485+c

c=3+485=15+485

c=12.6

So equation (2) becomes:

y=85x+12610

I opted for fractional form for consistency of format. This is because the 5 in 85 is prime. Thus division (convert to decimal) would introduce an error.

SagarStudy
Jan 16, 2016

y=−58x

If y=mx+c then m is called the slope of the line.

Here y=−58x+0

Therefore slope of the given line is −58=m1(Say).

If two lines are perpendicular then the product of their slopes is −1.

Let the slope of the line perpendicular to the given line be m2.

Then by definition m1⋅m2=−1.

⇒m2=−1m1=−1−58=85⇒m2=85

This is the slope of required line and the line required line also passes through (−6,3).

Using point slope form

y−y1=m2(x−x1)

⇒y−3=85(x−(−6))

⇒y−3=85(x+6)

⇒5y−15=8x+48

⇒8x−5y+63=0

This is the required line.

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