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9.Straight Line
hard
Co-ordinates of the orthocentre of the triangle whose vertices are $A(0, 0) , B(3, 4)$ and $C(4, 0)$ is
A
$(3, 1)$
B
$(3, 4)$
C
$(3, 3)$
D
$\left( {3,\frac{3}{4}\,} \right)$
Solution
To find orthocentre of the triangle formed by (0,0),(3,4) and (4,0)
Let $H$ be the orthocenter of $\triangle O A B$
Therefore (slope of $O P$ i.e $O H) .($ slope of $B A)=-1$ $\Rightarrow\left(\frac{y-0}{3-0}\right) \cdot\left(\frac{4-0}{3-4}\right)=-1 \Rightarrow-\frac{4}{3} y=-1 \Rightarrow y=\frac{3}{4}$
Therefore required orthocenter $=(3, y)=\left(3, \frac{3}{4}\right)$
Standard 11
Mathematics
