First slide

Cartesian plane

Question

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:

Moderate

A

$(2, - 1 / 2)$
B

$(1, 3 / 4)$
C

$(1, - 3 / 4)$
D

$(2, 1 / 2)$

Solution

Let $A B C$ be the triangle the coordinates of whose vertices are ; $4 (k, - 3 k), B, (5, k)$ and $C (- k, 2)$ . It is given that Area of $∆ A B C = 28$ sq. units
$= \frac{1}{2} |\begin{array}{rrr} k - 3 k 1 \\ 5 k 1 \\ - k 2 1 \end{array}| = \pm 28$
$\Rightarrow \frac{1}{2} |\begin{array}{rrr} k - 3 k 1 \\ 5 - k 4 k 0 \\ - 2 k 2 + 3 k 0 \end{array}| = \pm 28$
$\begin{array}{l} = (5 - k) (2 + 3 k) + 8 k^{2} = \pm 56 \\ = 5 k^{2} + 13 k + 66 = 0 or 5 k^{2} + 13 k - 46 = 0 \end{array}$
$\Rightarrow 5 k^{2} + 13 k - 46 = 0$ $[\begin{matrix} ∵ 5 k^{2} + 13 k + 66 = 0 \\ has imaginary roots \end{matrix}]$
$\Rightarrow (k - 2) (5 k + 23) = 0 \Rightarrow k = 2 [∵ k \in Z]$
Hence, the coordinates of vertices are $A (2, - 6), B (5, 2)$ and
$(- 2, 2)$
The equations of altitudes through vertices $A$ . and $C$ are .: $x = 2$ and $3 x + B y - 10 = 0$ respectively. These two altitudes intersect : $(2, 1 / 2) .$ Hence, the coordinates of the orthocentre are $(2, 1 / 2) .$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App