Write Equation in slope-intercept form of the line satis..

[LMande]

Write Equation in slope-intercept form of the line satisfying the given conditions.

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Okay guys... I'm at a complete lost with this question!

The line passes trhough (-5, 6) and is perpendicular to the line that has an x-intercept of 3 and a y-intercept of -9.

How the??? What the?? This is definitely a different language! Help on this question would be greatly appreciated!

 

[kinerry]

ok first off, do you know how to find the slope?

 

[galactus]

Re: Write Equation in slope-intercept form of the line satis

fying the given conditions.

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Okay guys... I'm at a complete lost with this question!

The line passes trhough (-5,6) and is perpendicular to the line that has an x-intercept of 3 and a y intercept of -9.

How the??? What the?? This is definitely a different language! Help on this question would be greatly appreciated!

The line given has x intercept of 3. That means its coordinates are (3,0)

Its y-intercept is -9. That means its coordinates are (0,-9)

So, you have a line with coordinates (3,0) and (0,-9). Find the equation of this line. You have the necessary info. Then use the negative reciprocal of its slope to find the slope of the line you're hunting for.

You then have slope and a point it passes through, (-5,6). Use y=mx+b to find its equation.

 

[LMande]

Slope

The slope m = (y₂ - y₁)/(x₂ - x₁)

Right?
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Edited by stapel -- Reason for edit: clarity of formatting.

 

[jonboy]

Hello LMande

Write equation for: The line passes through (-5, 6) and is perpendicular to the line that has an x-intercept of 3 and a y-intercept of -9

I am sure you meant this for finding slope: \(\displaystyle \L \;\frac{{\Delta y}}{{\Delta x}}=\frac{y_1-y_2}{x_1-x_2}\)

Like galactus said we have the points: \(\displaystyle \L \; (3,0)\;&\;(0,-9)\)

So: \(\displaystyle \L \;m=\frac{3-0}{0-(-9)}\;\Rightarrow\;m=\frac{1}{3}\;\Rightarrow\; \bot m=-3\)

So:\(\displaystyle \L \;y= -3x+b\)

Fill in one our coordinate: \(\displaystyle \L \;(-5, 6)\;\) to find constant \(\displaystyle \L \;b\)

Hence:\(\displaystyle \L \;6= -3(-5)+b\;\to\;-9=b\)

So our perpendicular equation is: \(\displaystyle \L \;y=-3x-9\)

Check: \(\displaystyle \L \;6=-3(-5)-9\;\to\; 6=6\)

Cool enough?