SATHEE: Straight Line and Pair of Straight Lines 1 Question 19

Straight Line and Pair of Straight Lines 1 Question 19

19.

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

(2017 Main)

(a) $(2, - \frac{1}{2})$

(b) $(1, \frac{3}{4})$

(d) $(2, \frac{1}{2})$

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Answer:

Correct Answer: 19. (d)

Solution:

Given, vertices of triangle are $(k, - 3 k), (5, k)$ and $(- k, 2)$ .

$\begin{array}{lr} ∴ & \frac{1}{2} | \begin{array}{ccc} k & - 3 k & 1 \\ 5 & k & 1 \\ - k & 2 & 1 \end{array} | = \pm 28 \\ \Rightarrow & | \begin{array}{ccc} k & - 3 k & 1 \\ 5 & k & 1 \\ - k & 2 & 1 \end{array} | = \pm 56 \\ \Rightarrow & k (k - 2) + 3 k (5 + k) + 1 (10 + k^{2}) = \pm 56 \\ \Rightarrow & k^{2} - 2 k + 15 k + 3 k^{2} + 10 + k^{2} = \pm 56 \\ \Rightarrow & 5 k^{2} + 13 k + 10 = \pm 56 \\ \Rightarrow & 5 k^{2} + 13 k - 66 = 0 \\ or & 5 k^{2} + 13 k - 46 = 0 \\ \Rightarrow & k = 2 \end{array}$

Thus, the coordinates of vertices of triangle are $A (2, - 6), B (5, 2)$ and $C (- 2, 2)$ .

Now, equation of altitude from vertex $A$ is

$y - (- 6) = \frac{- 1}{(\frac{2 - 2}{- 2 - 5})} (x - 2) \Rightarrow x = 2$ …….(i)

Equation of altitude from vertex $C$ is

$y - 2 = \frac{- 1}{[\frac{2 - (- 6)}{5 - 2}]} [x - (- 2)]$

$\Rightarrow 3 x + 8 y - 10 = 0$ …….(ii)

On solving Eqs. (i) and (ii), we get $x = 2$ and $y = \frac{1}{2}$ .

$∴$ Orthocentre $= (2, \frac{1}{2})$

Straight Line and Pair of Straight Lines 1 Question 19

19.

Answer:

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Straight Line and Pair of Straight Lines 1 Question 19

19.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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