Question

Find the orthocentre of the triangle whose vertices are

(0, 0), (6, 1)

and

(2, 3)

Accepted Answer

We have $A = (0, 0) B = (6, 1) & C = (2, 3)$

We need to find orthocentre of $△ A B C$ .

As we know that orthocentre is the intersecting point for all altitudes of the triangle .

Altitudes are nothing but perpendicular line (AE,BF and CD) from one side of triangle (either CA,AB or BC) to the opposite vertex.

Slope of BC= $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ = $m_{B C}$ = $\frac{3 - 1}{2 - 6} = \frac{- 1}{2}$

Slope of AE =perpendicular slope of BC

$m_{A E} \times m_{B C = - 1} m_{A E =} \frac{- 1}{m_{B C}} = \frac{- 1}{\frac{- 1}{2}} = 2$

The equation of AE is given as

$y - y_{1} = m (x - x_{1}) A = (0, 0) y = 2 x (1)$

Slope of AB

$m_{A B} = \frac{1 - 0}{6 - 0} = \frac{1}{6} m_{A B} \times m_{C D} = - 1 m_{C D} = \frac{- 1}{m_{A B}} = - 6$

Equation of CD is given as

$C = (2, 3) y - y_{1} = m (x - x_{1}) y - 3 = - 6 (x - 2) y = - 6 x + 15 6 x + y = 15 p u t t i n g y = 2 x f r o m (1) 6 x + 2 x = 15 8 x = 15 x = \frac{15}{8} y = \frac{15}{4}$

Hence the required orthocentre of triangle ABC is $O (\frac{15}{8}, \frac{15}{4})$

Find the orthocentre of the triangle whose vertices are $(0, 0), (6, 1)$ and $(2, 3)$

Important Questions on Straight Lines

Learn and Prepare for any exam you want !

Find the orthocentre of the triangle whose vertices are 0,0, 6,1 and 2,3

Important Questions on Straight Lines

Learn and Prepare for any exam you want !

Find the orthocentre of the triangle whose vertices are $(0, 0), (6, 1)$ and $(2, 3)$