Question

I: A straight line is such that the algebraic sum of the distance from any no. of fixed points is zero. Then that line always passes through a fixed point
II: The base of the triangle lie along the line $x=a$ and is of length $a.$If the area of the triangle is $a^2$ then the third vertex lies on $x=-a$ or $x=3a$.

Then which of the following is true.

• A. only I
• B. only II
• C. both I & II
• D. neither I nor II

Answer 1

The correct option is C both I & II

$I.$ Let $A(x_1,y_1)B(x_2,y_2)$ and line $ax+by+c=0$

$d_1=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$ and $d_2=\left|\frac{a{x}_2+b{y}_2+c}{\sqrt{a^2+b^2}}\right|$

$d_1+d_2=0\Rightarrow d_1+d_2=\frac{a(x_1+x_2)+b(y_1+y_2)+2c}{\sqrt{a^2+b^2}}=0$

$a(x_1+x_2)+b(y_1+y_2)+2c=0$

This line passes through some fixed point

$x=\frac{x_1+x_2}{2}$ and
$y=\frac{y_1+y_2}{2}$

So, I is True

$II.$ Given base length $=a$

$\therefore$ area$=\frac{1}{2}\times a\times$ height $=a^2$

Height $=2a$

So, another vertex lie on $x=-a$ or $x=3a$

Both I and II are True