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Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999 - 2 Marks)
- a)3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0
- b)3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0
- c)3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0
- d)3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Let PQR be a right angled isosceles triangle, right angled at P (2, 1)...
Let m be the slope of PQ then
⇒ m + 2 = 1 – 2m or – 1 + 2m = m + 2
⇒ m = – 1/3 or m = 3
⇒ m = – 1/3 or m = 3
As PR also makes ∠ 45° with RQ.
∴ The above two values of m are for PQ and PR.
∴ The above two values of m are for PQ and PR.
∴ Equation of PQ, 
⇒ 3y – 3 = – x + 2 ⇒ x + 3y – 5 = 0
and equation of PR is ⇒ 3x – y – 5 = 0
∴ Combined equation of PQ and PR is (x – 3y – 5) (3x – y – 5) = 0
⇒ 3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0
∴ Combined equation of PQ and PR is (x – 3y – 5) (3x – y – 5) = 0
⇒ 3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0
Most Upvoted Answer
Let PQR be a right angled isosceles triangle, right angled at P (2, 1)...
To find the equation of the line PQ, we need to find the slope of QR and then take its negative reciprocal.
The equation of line QR is given as 2x - y = 3. We can rewrite this equation in slope-intercept form by solving for y:
-y = -2x + 3
y = 2x - 3
The slope of line QR is 2.
Since PQ is perpendicular to QR, the slope of PQ is the negative reciprocal of the slope of QR. So the slope of PQ is -1/2.
Now we need to find the equation of line PQ using the point-slope form, y - y1 = m(x - x1), where (x1, y1) is a point on the line.
We have the point P(2, 1), so we can substitute these values into the equation:
y - 1 = -1/2(x - 2)
y - 1 = -1/2x + 1
y = -1/2x + 2
The equation of line PQ is y = -1/2x + 2.
To find the equation of line PR, we can use the fact that the sum of slopes of perpendicular lines is -1.
The slope of PQ is -1/2, so the slope of PR is 2.
We also have the point P(2, 1), so we can use the point-slope form again:
y - 1 = 2(x - 2)
y - 1 = 2x - 4
y = 2x - 3
The equation of line PR is y = 2x - 3.
Therefore, the equation representing the pair of lines PQ and PR is:
3x^2 + 4xy - y^2 + 7x - 2y + 2 = 0
So the answer is (a) 3x^2 + 4xy - y^2 + 7x - 2y + 2.
The equation of line QR is given as 2x - y = 3. We can rewrite this equation in slope-intercept form by solving for y:
-y = -2x + 3
y = 2x - 3
The slope of line QR is 2.
Since PQ is perpendicular to QR, the slope of PQ is the negative reciprocal of the slope of QR. So the slope of PQ is -1/2.
Now we need to find the equation of line PQ using the point-slope form, y - y1 = m(x - x1), where (x1, y1) is a point on the line.
We have the point P(2, 1), so we can substitute these values into the equation:
y - 1 = -1/2(x - 2)
y - 1 = -1/2x + 1
y = -1/2x + 2
The equation of line PQ is y = -1/2x + 2.
To find the equation of line PR, we can use the fact that the sum of slopes of perpendicular lines is -1.
The slope of PQ is -1/2, so the slope of PR is 2.
We also have the point P(2, 1), so we can use the point-slope form again:
y - 1 = 2(x - 2)
y - 1 = 2x - 4
y = 2x - 3
The equation of line PR is y = 2x - 3.
Therefore, the equation representing the pair of lines PQ and PR is:
3x^2 + 4xy - y^2 + 7x - 2y + 2 = 0
So the answer is (a) 3x^2 + 4xy - y^2 + 7x - 2y + 2.
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Let PQR be a right angled isosceles triangle, right angled at P (2, 1)...
The correct answer is option.B)
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Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999 - 2 Marks)a)3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0b)3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0c)3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0d)3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0Correct answer is option 'B'. Can you explain this answer?
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Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999 - 2 Marks)a)3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0b)3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0c)3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0d)3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999 - 2 Marks)a)3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0b)3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0c)3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0d)3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999 - 2 Marks)a)3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0b)3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0c)3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0d)3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0Correct answer is option 'B'. Can you explain this answer?.
Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999 - 2 Marks)a)3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0b)3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0c)3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0d)3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999 - 2 Marks)a)3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0b)3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0c)3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0d)3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999 - 2 Marks)a)3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0b)3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0c)3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0d)3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0Correct answer is option 'B'. Can you explain this answer?.
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Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999 - 2 Marks)a)3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0b)3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0c)3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0d)3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999 - 2 Marks)a)3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0b)3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0c)3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0d)3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999 - 2 Marks)a)3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0b)3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0c)3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0d)3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999 - 2 Marks)a)3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0b)3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0c)3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0d)3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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