(person who helps gets medal) Find the equation of a line that passes through the point (4,2) that is perpendicular to the line y = x. Show your work.
\[y = \frac{ 4 }{ 3 } X\]
is that your answer?
no the fraction did not appear when I copy and paste
the 4/3 is suppose to be in the question
so what would the slope be of a line that is perpendicular to y=3/4x?
How do we find that?
mmm im having trouble since its a fraction ~
do i put it like this y2 - y1 -------- x2 - x1
If one line has a slope of \(\frac{1}{2}\), then the line that is perpendicular to it will be the negative reciprocal \(-2\)
nope to find the slope of a line that is perpendicular to that, you just do the opposite reciprocal of the slope of y=3/4x
so i make it negative ? 3 / 4 - ?
and reciprocal
??
@Vocaloid
Yes, so now we have our new slope. Now we must plug in the other values that we're given. We're told that the line runs through the point (4, 2) Equation of a Line: \[\large (y-y_1)=m(x-x_1)\]
Plug in the point into \(y_1\) and \(x_1\) and our slope \(m\) that we just also found.
\[(y - 2) = -\frac{ 3 }{ 4 }( x - 4)\]
this ?
Yep! X) Now simplify. So distribute the (-3/4) and then add 2 to both sides.
how would u distribute the -3/4 ?
@CShrix
Let's look at 2(x+1) By distributing, we multiply each term by 2 so 2(x+1) = 2x+2
sooo you multiply the fraction by x and 4 ?
could you draw it out instead of saying it? i can under stand it so much better
|dw:1447730905953:dw| We have to multiply x and (-4) by (-3/4)
so it would be -3/4x - 12/3 ?
?
(3/4)(4) is not (12/3)
what is it then ~
im super bad at fractions
\[y-2=-\frac{3}{4}x+\frac{12}{4}\]
thats the answer ?
Actually it depends on which form they want. If they want it in `slope-intercept `form, then we must add both sides by 2 to isolate y. But if they want the answer in `point-slope` form, then our answer is just \(\large (y - 2) = -\frac{ 3 }{ 4 }( x - 4)\) If it doesn't specify, then I would just leave it in point-slope form, which we already have.
ok so yes alright thank you ! :)
You are welcome :)
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