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…
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 triangle formed by the lines $x=0, y=0$ and $x+y=1$ is
5
$\sqrt{5}$
$\sqrt{20}$
20
Solution
$\begin{gathered} 7 x+4 y=15 \\
7 x+7 y+10=0 \end{gathered}$ distance between P and $\mathrm{B}=\sqrt{5}$
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