What is the equation of the line perpendicular to #y=-5/8x # that passes through # (-6,3) #?

2 Answers
Jan 16, 2016

Answer:

#y=8/5x+126/10#

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: #-1/m#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Find the generic equation of the line perpendicular to the original")#

Given equation: #y_1=-5/8x#...............................(1)

The equation perpendicular to this will be

#color(white)(xxxxxxxx)color(blue)(y_2=+8/5x+c)#......................................(2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("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:

#y_2=3=8/5(-6)+c#

#y_2=3=-48/5+c#

#c=3+48/5 = (15+48)/5#

#c=12.6#

So equation (2) becomes:

#y=8/5x+126/10#

I opted for fractional form for consistency of format. This is because the 5 in #8/5# is prime. Thus division (convert to decimal) would introduce an error.

Jan 16, 2016

#y=-5/8x#

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

Here #y=-5/8x+0#

Therefore slope of the given line is #-5/8=m_1 (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 #m_2#.

Then by definition #m_1*m_2=-1#.

#implies m_2=-1/m_1=-1/(-5/8)=8/5 implies m_2=8/5#

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

Using point slope form

#y-y_1=m_2(x-x_1)#

#implies y-3=8/5(x-(-6))#

#implies y-3=8/5(x+6)#

#implies 5y-15=8x+48#

#implies 8x-5y+63=0#

This is the required line.