A line passes through #(2,2)# and cuts a triangle of area #9 " units"^2# from the first quadrant. The sum of all possible values for the slope of such a line, is?

A) -2.5
B) -2
C) -1.5
D) -1

2 Answers
Shwetank Mauria
Mar 6, 2018

Sum of slopes is #-2.5# and answer is (A).

Explanation:

A line passing through point #(2,2)# and having a slope #m# has an equation

#y-2=m(x-2)#

or #mx-y=2m-2=2(m-1)#

or #(mx)/(2(m-1))-y/(2(m-1))=1#

or #x/((2(m-1))/m)+y/(-2(m-1))=1# (intercept form of equation)

Observe that such a line forming a triangle in #Q1#, will have a negative slope.

and its #x#-intercept is #(2(m-1))/m# and #y#-intercept is #-2(m-1)#

and hence area of triangle so formed is

#1/2xx(2(m-1))/mxx-2(m-1)# and as area is #9# we have

#-2(m-1)^2/m=9#

or #-2(m-1)^2=9m#

or #-2m^2+4m-2=9m#

or #2m^2+5m+2=0#

or #(2m+1)(m+2)=0#

i.e. #m=-1/2# or #m=-2#

Hence sum of slopes is #-2.5# i.e. answer is (A)

and lines are #x+2y=6# or #2x+y=6#

graph{(x+2y-6)(2x+y-6)=0 [-6.69, 13.31, -2.62, 7.38]}

Ratnaker Mehta
Mar 6, 2018

# A) -2.5#.

Explanation:

Let, the X-intercept and Y-intercept of the line, say #l#, through the

point #P(2,2)# be #a and b#, resp.

Clearly, # l : x/a+y/b=1#.

#P in l"...[Given] "rArr 2/a+2/b=1.........(star)#.

Observe that, #l# makes, with the Axes, such a right-triangle,

of which, the lengths of the sides making the right angle are

#a and b#.

Hence, the area of the right-triangle is #1/2ab#.

Knowing that, #1/2ab=9, b=18/a.#

Then, #(star) rArr 2/a+2*a/18=1, or, a^2-9a+18=0#.

#rArr a=6, or 3, & :., b=3, or 6#.

These give #2# values of slope #m=-b/a# of #l#, namely,

#m_1=-1/2, or m_2=-2," giving the sum of slopes, "-2.5#.

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