*k*such that the line

*3x + ky = 17*is perpendicular to the line

*8x - 5y = 26*

Find *k* such that the line *3x + ky = 17 *is perpendicular to the line
*8x - 5y = 26*

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Hi Ryan;

Both formulas are in Standard Format...

Ax+By=C A>0, B>0

We need Slope-Intercept Format...

8x - 5y = 26

Let's subtract 8x from both sides...

-8x+8x-5y=-8x+26

-5y=-8x+26

Let's divide both sides by -5...

(-5y)/-5=[(-8/-5)x]+(26/-5)

y=(8/5)x+(-26/5)

The slope is 8/5

The perpendicular slope is -5/8.

3x + ky = 17

Let's subtract 3x from both sides...

-3x+3x+ky=-3x+17

ky=-3x+17

Let's divide both sides by k...

(ky)/k=[(-3/k)x]+(17/k)

y=(-3/k)x+(17/k)

-3/k=-5/8

Let's flip...

-k/3=-8/5

Let's multiply both sides by -3...

-3(-k/3)=(-8/5)(-3)

k=24/5

Lines a_{1}x+b_{1}y+c_{1}=0 and a_{2}x+b_{2}y+c_{2}=0 are perpendicular if a_{1}a_{2}+b_{1}b_{2}=0.

In this case it means that 3*8-5k=0, from which it follows that k=24/5=4.8

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