Question

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

Answer 1

`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)

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`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.

Answer 2

`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.