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

Question

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

Accepted Answer

A is orthocentre of above $\Delta$.

$$\begin{aligned}
&O \text { is orthocentre of above } \Delta \text {. }\
&O A=\sqrt{5}
\end{aligned}$$

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