Question

orthocentre of the triangle with vertices (0,0) (3,4) (4,0) is ​

Answer 1

Answer:

(0,0) is your answer for that question

Answer 2

Step-by-step explanation:

We are given that 3 vertices of traingle (0,0) (3,4) (4,0)

To find : Orthocentre

Orthocentre is the point at which any two Altitudes of the traingle meets

So, By figure the(refer attachment) (point H) will be the orthocentre.

As it lie on the intersecting points of the two attitudes

Let us assume the point H be the orthocentre of ∆OAB

Clearly its altitude will be (3,y)

•°• (slope of OP that is OH) × (slope of BA) = -1

[°•° As we know the product of any two perpendicular lines is - 1 ]

Slope formula = $\sf{\frac{y_2 - y_1}{x_1 - x_2} }$

$\implies{\sf{ ( \frac{y - 0}{3 - 0} ) \times ( \frac{4 - 0}{3 - 4} ) = - 1 }}$

$\\ \\ \implies{\sf{( \frac{y}{3}) \times ( \frac{4}{ - 1} ) = - 1 }}$

$\\ \\ \implies{\sf{ - \frac{4}{3} y = - 1}}$

$\\ \\ \implies{\mathsf{ \frac{4}{3}y = 1 }}$

$\\ \\ \implies{\sf{y = \frac{3}{4} }}$

Thus, Required orthocentre is (3,y) =$\boxed{\mathsf{(3, \frac{3}{4} )}}$