Question

The orthocentre of the triangle having vertices at $(2,3)(2,5)(4,3)$ is

• A. $(0,0)$
• B. $(4,3)$
• C. $(2,3)$
• D. $none$

Answer 1

The correct option is C $(2,3)$

We have,

Given point are

$(2,3),(2,5),(4,3)$

So,

Let

$A(x_1,y_1)=(2,3)$

$B(x_2,y_2)=(2,5)$

$C(x_3,y_3)=(4,3)$

$\text{Slope}\text{of}\text{AC}\text{=}\frac{y_3-y_1}{x_3-x_1}=\frac{3-3}{4-2}=0$

Let the slope of line $BD$ $=$ m

$\text{So,}\text{BD}\perp AC$

$m\times 0=1$

$m=\infty$

So, equation of $BD\Rightarrow y-5=\infty (x-2)$

$\Rightarrow x-2=0......\left(1\right)$

$x=2$

Slope of line

$AB\Rightarrow \frac{y_2-y_1}{n_2-n_1}=\frac{5-3}{2-2}=\infty$

$CE\perp AB$

Let the slope of CE be n.

$n\times \infty =-1,n=0$

Now,

Equation of CE is

$y-3=n(x-4)$

$y-3=0(n-4)$

$y-3=0......\left(2\right)$

From equation (1) and (2) to, and

We get.

$(n,y)=(2,3)(\text{othocentre}\text{.)}$

Hence, this is the answer.