Let the three sides of a triangle be on the lines 4x-7y+10=0, x+y=5 and 7x+4y=15.

Question

Let the three sides of a triangle be on the lines $4x-7y+10=0,\ x+y=5$ and $7x+4y=15$. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines $x=0,\ y=0$ and $x+y=1$ is:

Accepted Answer

The three given lines are mutually perpendicular in pairs appropriately, such that the orthocentre of $\triangle ABC$ (formed by $4x-7y+10=0,\ x+y=5,\ 7x+4y=15$) turns out to be $A=(0,0)$ --- the same as the vertex where the perpendicular pair of sides meet (the source identifies this directly by inspection of slopes).

For the second triangle (formed by $x=0,y=0,x+y=1$, a right triangle with the right angle at the origin), the orthocentre is the origin $O=(0,0)$.

The orthocentre of the second triangle is $O$, and $OA=\sqrt5$ is the distance to the orthocentre of the first triangle (which lies at a point identified from the perpendicularity conditions as being at distance $\sqrt5$ from $O$).

$$\text{Required distance} = \sqrt5$$

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