SATHEE: Properties of Triangles 3 Question 2

Properties of Triangles 3 Question 2

2. Let the equations of two sides of a triangle be $3 x - 2 y + 6 = 0$ and $4 x + 5 y - 20 = 0$ . If the orthocentre of this triangle is at $(1, 1)$ then the equation of its third side is

(2019 Main, 9 Jan II)

(a) $122 y - 26 x - 1675 = 0$

(b) $26 x - 122 y - 1675 = 0$

(d) $26 x + 61 y + 1675 = 0$

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Answer:

Correct Answer: 2. (b)

Solution:

Let equation of $A B$ be $4 x + 5 y - 20 = 0$ and $A C$ be $3 x - 2 y + 6 = 0$

Clearly, slope of $A C = \frac{3}{2}$

$[∵ slope of a x + b y + c = 0 is - \frac{a}{b}]$

$∴$ Slope of altitude $B H$ , which is perpendicular to

$A C = - \frac{2}{3} \cdot ∵ m_{B H} = - \frac{1}{m_{A C}}$

Equation of $B H$ is given by $y - y_{1} = m (x - x_{1})$

Here, $m = - \frac{2}{3}, x_{1} = 1$ and $y_{1} = 1$

$∴ y - 1 = - \frac{2}{3} (x - 1)$

$\Rightarrow 2 x + 3 y - 5 = 0$

Now, equation of $A B$ is $4 x + 5 y - 20 = 0$ and equation of $B H$ is $2 x + 3 y - 5 = 0$

Solving these, we get point of intersection (i.e. coordinates of $B$ ).

$\begin{aligned} 4 x + 5 y - 20 = 0 \\ 4 x + 6 y - 10 = 0 \end{aligned} \Rightarrow y = - 10$

On substituting $y = - 10$ in $2 x + 3 y - 5 = 0$ , we get $x = \frac{35}{2}$

$∴ B \frac{35}{2}, - 10$

Solving $4 x + 5 y - 20 = 0$ and $3 x - 2 y + 6 = 0$ , we get coordinate of $A$ .

$Extra close brace or missing open brace$

Now, slope of $A H = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{\frac{84}{23} - 1}{\frac{10}{23} - 1} = \frac{61}{- 13}$ .

$∵ B C$ is perpendicular to $A H$ . $∴$ Slope of $B C$ is $\frac{13}{61}$

$∵ m_{B C} = - \frac{1}{m_{A H}}$

Now, equation of line $B C$ is given by $y - y_{1} = m (x - x_{1})$ , where $(x_{1}, y_{1})$ are coordinates of $B$ .

$\begin{array}{ll} ∴ & y - (- 10) = \frac{13}{61} x - \frac{35}{2} \\ \Rightarrow & y + 10 = \frac{13}{61 \times 2} (2 x - 35) \\ \Rightarrow & 122 y + 1220 = 26 x - 455 \\ \Rightarrow & 26 x - 122 y - 1675 = 0 \end{array}$

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Properties of Triangles 3 Question 2

2. Let the equations of two sides of a triangle be $3 x - 2 y + 6 = 0$ and $4 x + 5 y - 20 = 0$ . If the orthocentre of this triangle is at $(1, 1)$ then the equation of its third side is

Answer:

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Properties of Triangles 3 Question 2

2. Let the equations of two sides of a triangle be 3x−2y+6=0 and 4x+5y−20=0. If the orthocentre of this triangle is at (1,1) then the equation of its third side is

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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2. Let the equations of two sides of a triangle be $3 x - 2 y + 6 = 0$ and $4 x + 5 y - 20 = 0$ . If the orthocentre of this triangle is at $(1, 1)$ then the equation of its third side is