find the equation of the straight line which makes equal intercepts on the axes and passes through points (2,-7)
@hartnn please help
sure
:)
the equation of line can be y=mx+c do u know how to get the intercepts?
yeah! If we need to find y intercept then we'll take x=0
so find x and y intercepts for me.
for that I'll have to find m first How would I do that?
put x=0 in y=mx+c u get y=c<-----y-intercept. put y=0 in y=mx+c 0=mx+c mx=-c x= -c/m <----x-intercept
and?
but did u get that? now the line makes equal intercepts, so you have x intercept= y-intercept c= -c/m can u find m from here ?
yep m= -1
absolutely! (if you want you can remember that slope of line with equal intercepts is always -1, or you can do this exercise) so now your equation is y=-x+c ok?
yeah
u have point (x,y)=(2,-7) on it so just substitute x=2 and y=-7 in y=-x+c and find c, try it
I did that but what would be the equation?
what is c ?
-5
-7=-2+c c= -5, correct. u had slope m=-1, u got c= -5 just put it in y=mx+c
ohk and here too m= -1 right?
its the same line! yes, slope = m= -1
x + y + 5 =0
that is correct! did u understand everything?
yeah thanks a lot! You are an amazing tutor ;)
welcome ^_^
thanks for the compliment :)
:)
Refer to the attachment.
how did u get that graph @robtobey ,the equation and intercepts are incorrect there.
how is m=1 ?
in your graph, u have y-intercept=-9 and x-intercept=9 so the intercepts are only equal in magnitude, if the question intends to have intercepts of equal magnitude, the there will be two lines as solution, one yours y=x-9, and one what we have derived, x+y+5=0. since the question asks for only one line, we assume that both intercepts must be equal in magnitude and sign, which gives the equation as x+y+5=0 only.
Well the attachment below should do the trick, from the Mathematica program.
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