Equation of line in Straight lines

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Question

The distance between the circumcentre and the orthocentre of the triangle whose vertices are $(0, 0), (6, 8),$ and $(- 4, 3)$ is L. Then the value of $\frac{2}{\sqrt{5}} L$ is __

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Solution

Given vertices of triangle are $O (0, 0), B (6, 8)$ and $C (- 4, 3)$ .
Slope of $OB = \frac{8}{6} = \frac{4}{3}$
Slope of $OC = \frac{3}{4}$
$∴ ∠ BOC = \frac{π}{2}$
$ΔBOC$ is right -angled at O.
Circumcenter=Midpoint of hypotenuse $BC = (1, \frac{11}{2})$
Required distance $= \sqrt{(1 + \frac{121}{4})} = \frac{5 \sqrt{5}}{2} units$

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Equation of line in Straight lines

Remember concepts with our Masterclasses.

The distance between the circumcentre and the orthocentre of the triangle whose vertices are 0,0,6,8, and −4,3 is L. Then the value of 25L is __

Given vertices of triangle are O0,0, B6,8 and C−4,3. Slope of OB=86=43Slope of OC=34∴ ∠BOC=π2ΔBOC is right -angled at O.Circumcenter=Midpoint of hypotenuse BC=1,112Required distance =1+1214=552units

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The distance between the circumcentre and the orthocentre of the triangle whose vertices are $(0, 0), (6, 8),$ and $(- 4, 3)$ is L. Then the value of $\frac{2}{\sqrt{5}} L$ is __