Question

A line passes through $(2,2)$ and cuts a triangle of area $9squareunits$ from the first quadrant. The sum of all possible values for the slope of a line, is

• A. $-2.5$
• B. $-2$
• C. $-1.5$
• D. $-1$

Answer 1

Answer: (A)

$-2.5$

Let the slope of a line be =$m$ , and it pass through point $(2,2)$ then its equation is

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

$y=mx-2m+2$

Now, area of triangle in such a case is half the product of x-intercept and y-intercept.

x-intercept is given by putting y=0 i.e. mx−2m+2=0 or x$=$ $\frac{2m-2}{m}$ and

y-intercept is given by putting x=0 i.e. y=−2m+2.

or 18=−4m2+8m−4m

$\because$ Area of triangle $=\frac{1}{2}\times base\times hight$

$2A=(-2m+2)\times \frac{(2m-2)}{m}=\frac{-4m+8m-4}{m}$

$2Am=-4m^2+8m-4$

Put $A=9$ unit, we get

18m=−4m${}^{{}^2}$+8m−4

$4m^2+10m+4=0$

we know that,in a quadratic equation $a{x}^2+bx+c=0$,

sum of roots is $=-\frac{b}{a}$

sum of slopes is $-\frac{10}{4}=-2.5$