Let k be an integer such that the triangle with vertices (k$$, −$$3k),(5$$, $$k) and $$($$−k, $$2)$$ has area $$28$$ sq. units. Then the orthocentre of this triangle is at the point:

Question

Let k be an integer such that the triangle with vertices (k$$, −$$3k),(5$$, $$k) and $$($$−k, $$2)$$ has area $$28$$ sq. units. Then the orthocentre of this triangle is at the point:

Infinity Learn · Accepted Answer

Let ABC be the triangle the coordinates of whose vertices are ;$$4$$ (k$$, $$-3k), B, (5$$, $$k)  and  $$C(-k$$, $$2)$$. It is given that Area of∆ABC $$=$$ $$28$$ sq. units $$=12k$$    −$$3k$$    $$15$$    k    $$1$$−k    $$2$$    $$1=$$±$$28$$⇒ $$12k$$    −$$3k$$    $$15$$−k    $$4k$$    $$0$$−$$2k$$    $$2+3k$$    $$0=$$±$$28=(5$$−$$k)(2+3k)+8k2=$$±$$56=5k2+13k+66=0$$ or $$5k2+13k$$−$$46=0$$⇒$$5k2+13k$$−$$46=0$$     ∵$$5k2+13k+66=0$$ has imaginary roots ⇒(k$$−$$2)(5k+23)=0$$⇒$$k=2$$ [∵k∈Z]Hence, the coordinates of vertices are A $$($$ $$2$$, $$-$$ $$6), B (5$$, $$2) $$and($$−$$2$$, $$2)The$$ equations of altitudes through vertices A. andC are .:$$x=$$ $$2$$ and $$3x$$ $$+$$ By $$-10$$ $$=$$ $$0$$ respectively. These two altitudes intersect : (2$$, $$1/2). Hence, the coordinates of the orthocentre are $$($$ $$2$$, $$1/$$ $$2)$$.

Let $k$ be an integer such that the triangle with vertices $(k, - 3 k), (5, k) and (- k, 2)$ has area 28 sq. units. Then the orthocentre of this triangle is at the point:

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Let k be an integer such that the triangle with vertices (k, −3k),(5, k) and (−k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point:

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Let $k$ be an integer such that the triangle with vertices $(k, - 3 k), (5, k) and (- k, 2)$ has area 28 sq. units. Then the orthocentre of this triangle is at the point: