JEE MAIN - Mathematics (2025 - 4th April Morning Shift - No. 18)
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
$\sqrt{20}$
$20$
$\sqrt{5}$
$5$
Explanation
_4th_April_Morning_Shift_en_18_1.png)
A is orthocentre of above $\Delta$.
_4th_April_Morning_Shift_en_18_2.png)
$$\begin{aligned} &O \text { is orthocentre of above } \Delta \text {. }\\ &O A=\sqrt{5} \end{aligned}$$
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