# Find the coordinates of point A.

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.

• 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. Commented Feb 18, 2022 at 19:16
• You wrote ℎ=−133, is there any possibility you may have miscalculated it instead of $h=-\frac{11}{3}$
– by24
Commented Feb 18, 2022 at 19:23

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

$$y_A = \pm 7$$