[SPLIT] Find the equation of the line through (6, 2) and....

[dieggo]

Could you explain me how to do this problem?

Find the equation of a line that goes through (6,2) and is perpendicular to the line with x-intercept (2,0) and y-intercept (0,-3). Write the equation in standard form.

Thank you!

 

[jonboy]

Could you explain me how to do this problem?

Find the equation of a line that goes through (6,2) and is perpendicular to the line with x-intercept (2,0) and y-intercept (0,-3). Write the equation in standard form.

Good Morning!

Ok first lets find the equation of the line with points of (2,0) and (0,-3) so we can determine the perpendicular (negative reciprocal) slope and the equation of the line perpendicular to it. We will call it equation [1]

First thing to do is find that slope: \(\displaystyle \L \;m\,=\,\frac{\Delta y}{\Delta x}\,=\,\frac{0-(-3)}{2-0}\,\to\,\frac{3}{2}\)

So we now have: \(\displaystyle \L \;y\,=\,\frac{3}{2}x\,+\,b\)

Since lines are consistent in they way the x's and y's change, just plug in one of our point to find \(\displaystyle b\) (y-intercept) .

\(\displaystyle \L \;0\,=\,\frac{3}{2}(2)\,+\,b\)

\(\displaystyle \L \;0\,=\,3\,+\,b\)

\(\displaystyle \L \,-3\,=\,b\)

................So equation [1]'s equation is \(\displaystyle \L \;y\,=\,\frac{3}{2}x\,-3\)

So as we know a perpendicular line has a negative reciprocal slope. You just flip the fraction and then change the sign.

So \(\displaystyle \L \bot m\,=\,-\,\frac{2}{3}\)

Now we have: \(\displaystyle \L \;y\,=\,-\,\frac{2}{3}x\,+\,b\)

Fill in the coordinate we know:\(\displaystyle \L \;\,2=\,-\,\frac{2}{3}(6)\,+\,b\)

\(\displaystyle \L \;2\,=\,-\,4\,+\,b\)

\(\displaystyle \L 6\,=\,b\)

..................So our perpendicular equation to equation [1]:\(\displaystyle \L \;y\,=\,-\,\frac{2}{3}x\,+\,6\;\to\;\frac{2}{3}x\,+\,y\,=\,6\)


If you want you can multiply \(\displaystyle \frac{2}{3}x\,+\,y\,=\,6\) by \(\displaystyle 3\) to get rid of the fraction.

 

[wantedcriminal03]

Re: please help me, pleaseee

\(\displaystyle \L y_2\,-\,y_1\,=\,m(x_2\,-\,x_1)\)

\(\displaystyle \L -3\,-\,0\,=\,m(0\,-\,2)\)

\(\displaystyle \L \frac{-3}{-2}\\)=m=\(\displaystyle \L \frac{3}{2}\\)

Gradient of line perpendicular is the negative reciprocal of \(\displaystyle \L \frac{3}{2}\\) = \(\displaystyle \L-\ \frac{2}{3}\\)

\(\displaystyle \L y-2\,=\,-\, \frac{2}{3}\(x-6)\)

\(\displaystyle \L y-2\,=\,-\, \frac{2}{3}x\,+4\)

\(\displaystyle \L y\,=\,-\, \frac{2}{3}x\,+6\)

 

[jonboy]

Re: please help me, pleaseee

\(\displaystyle \L y\,=\, \frac{-2x}{3}\ +6\)

We match. I was going to put it in just slope intercept form till I saw that the queations requires "standard" form.