Question

Consider the lines represented by the equation $(x^2+xy-x)(x-y)=0$, forming a triangle, then
$\begin{array}{cc}Column1& Column2\\ \left(a\right)Orthocentreoftraingle& (\frac{1}{6},\frac{1}{2})\\ \left(b\right)Circumcenter& \left(\frac{1}{2+2\sqrt{2}}\right)\\ \left(c\right)Centroid& (0,\frac{1}{2})\\ \left(d\right)Incenter& (\frac{1}{2},\frac{1}{2})\end{array}$

• A. a-r, b-s, c-p, d-q
• B. b-r, b-s, c-q, d-p
• C. a-s, b-r, c-q, d-p1
• D. a-s, b-r, c-p, d-q

Answer 1

The correct option is D

a-s, b-r, c-p, d-q

Given lines are $(x^2+xy-x)(x-y)=0$
x(x+y-1)(x-y)=0
or x=0, x+y-1=0 and x-y=0 forms triangle OAB as shown in the diagram.

<> Intersection points these line A(0,1), $B(\frac{1}{2},\frac{1}{2})$, and O(0,0)
Triangle is right angled at point B, hence
Orthocentre is point $B(\frac{1}{2},\frac{1}{2})$,
Also, circumcentre is the mid-point of OA which is $\frac{(0+0)}{2},\frac{(0+1)}{2}=(0,\frac{1}{2})$

Centroid is $(\frac{0+\frac{1}{2}+0}{3},\frac{0+\frac{1}{2}+1}{3})or(\frac{1}{6},\frac{1}{2})$
Also OA=b=1
OB=OC=$\frac{1}{\sqrt{2}}$
Incenter is $(\frac{0\times \frac{1}{\sqrt{2}}+\frac{1}{2}\times 1+0\times \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}+1+\frac{1}{\sqrt{2}}},\frac{0\times \frac{1}{\sqrt{2}}+\frac{1}{2}\times 1+1\times \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}+1+\frac{1}{\sqrt{2}}})=(\frac{1}{2+2\sqrt{2}},\frac{1}{2})$

So, a-s, b-r, c-p, d-q