2
$\begingroup$

The Problem -

In $\triangle ABC,$ $B=(-7,1),C=(-5,-5)$. Centroid $G=(h,k)$ and orthocentre $H=(3h,3k)$. If $A=(1,Y_A)$ then find $Y_A$.

First I found that $h=-\dfrac{13}{3}⇒ 3h=-13⇒ H\equiv(-13,k).$

Then I found that $k=\dfrac{Y_A-4}{4}⇒ 3k=Y_A-4⇒ H\equiv(-13,Y_A-4).$

Then we get Slope $BC=\dfrac{1+5}{-7+5}=\dfrac{6}{-2}=-3⇒ $Slope of altitude from $A=\dfrac{1}{3}$ .

Like what do I do from here? If I substitute value of $H$ in equation of altitude I just get $0=0$.I am stuck and cannot proceed further. Can someone please help? Thanks.

$\endgroup$
2
  • $\begingroup$ Since this is a Geometry problem, please edit your question by imbedding a diagram directly into the question, as opposed to providing a link to an external (uploaded) file. Typically, any reasonably sized jpeg will work. See the Images section, at the bottom of this article. See also, this article. $\endgroup$ Commented Feb 18, 2022 at 19:16
  • $\begingroup$ You wrote ℎ=−133, is there any possibility you may have miscalculated it instead of $h=-\frac{11}{3}$ $\endgroup$
    – by24
    Commented Feb 18, 2022 at 19:23

1 Answer 1

1
$\begingroup$

Coordinates of $G$ should be $ \displaystyle \left(\frac{- 7 - 5 + 1}{3}, \frac{1 - 5 + y_A}{3}\right)$

That gives $ \displaystyle h = - \frac{11}{3}, y_A - 4 = 3k$

and so, we have coordinates of $ \displaystyle H \text { as } \left( - 11, y_A - 4 \right)$

Slope of $ \displaystyle AC, ~m_1 = \frac{y_A + 5}{6}$,

Slope of $ \displaystyle BH, ~m_2 = \frac{y_A - 5}{-4}$,

As $BH \perp AC, ~m_1 \cdot m_2 = -1 \implies y_A^2 - 25 = 24$

$y_A = \pm 7$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .