Question

Consider the triangle ABC having vertex A (1, 1) and its orthocentre is (2, 4). Also side AB & BC are members of the family of line, ax + by + c = 0 where 2b = a + c

Answer 1

We have,

Side $AB$ and BC are the members of the family of line

$ax+by+c=0......\left(1\right)$

And $2b=a+c$

As the passing through the point $(1,1)$.

Then,

$a+b+c=0$

As a,b,c are in arithmetic progression,

Hence, if d is common difference,

Then,

$b=b$

$c=b+d$

$a=b-d$

Hence,

$a+b+c=0$

If $b=0$

So,

$a+b+c=0$

$a+c=0$

$\Rightarrow a=-c$

Hence, $ax+by+c=0$

Becomes

$ax+0y-a=0$

$\Rightarrow ax-a=0$

$\Rightarrow ax=a$

$\Rightarrow x=1$

Then consider point C. Because AB is on line equation (1) and the orthocenter is a.

$(2,4)$

Hence, C must be of type $(q,4)$

BC lies on the line

$ax+by+c=0$

Hence, put

$B(1,p)$

$\Rightarrow a+bp+c=0$

Also, $a+c=2b$

As

$2b+bp=0$

$p=-2$

Hence, the coordinate of B are $B(1,-2)$

The slope of line through B and C is $\frac{4+2}{q-1}=\frac{6}{q-1}$

The line perpendicular to line BC passing through A

Then,

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

$\Rightarrow y-1=\frac{-\frac{1}{6}}{q-1}(x-1)$

$\Rightarrow y-1=\frac{1-q}{6}(x-1)$

This line must passes through orthocenter $(2,4)$

$(x,y)=(2,4)$ put here,

$4-1=\frac{1-q}{6}(2-1)$

$\Rightarrow 3=\frac{1-q}{6}\times 1$

$\Rightarrow 18=1-q$

$\Rightarrow q=-17$

Hence, the coordinates of C are $(-17,4)$.

Hence, this is the answer.