Question

Let the algebraic sum of the perpendicular distance from the points (2,0), (0,2) and (1,1) to a variable straight line be zero; then the line passes through a fixed point whose coordinates are:

• A. $(1,1)$
• B. $(\frac{1}{2},\frac{1}{2})$
• C. $(\frac{3}{2},\frac{3}{2})$
• D. centroid of triangle

Answer 1

The correct option is A $(1,1)$

We have,

$(x_1,y_1)=(2,0)$

$(x_2,y_2)=(0,2)$

$(x_3,y_3)=(1,1)$

Then,

We know that,

Equation of line

$ax+by+c=0$

Distance from point (2,0)

$D_1=\frac{(2a+0b+c)}{k}......\left(1\right)$

Distance from point (0,2)

$D_2=\frac{(0a+2b+c)}{k}......\left(2\right)$

Distance from point (1,1)

$D_3=\frac{(a+b+c)}{k}......\left(3\right)$

Where $k=\sqrt{a^2+b^2}$

Now, on adding (1)+(2)+(3) to, and we get,

$D_1+D_2+D_3=\frac{3a+3b+3c}{k}$

$D_1+D_2+D_3=\frac{3(a+b+c)}{k}$

As the sum is zero.

Hence,

$a+b+c=0$

From equation of line,

$ax+by+c=0$

Taking $(x_1,y_1)=(1,1)$

$a+b+c=0$

Hence, this is the answer.