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For the triangle whose sides are along the lines y = 15, 3x – 4y = 0, 5x + 12y = 0, the incentre is :
- a)(1, 8)
- b)(8, 1)
- c)(-1, 8)
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
For the triangle whose sides are along the lines y = 15, 3x – 4y...
Given equations:
3x – 4y = 0 …(1)
5x+12y = 0 …(2)
Y-15 = 0 …(3)
From the given equations, (1), (2) and (3) represent the sides AB, BC and CA respectively.
Solving (1) and (2), we get
x= 0, and y= 0
Therefore, the side AB and BC intersect at the point B (0, 0)
Solving (1) and (3), we get
x= 20, y= 15
Hence, the side AB and CA intersect at the point A (20, 15)
Solving (2) and (3), we get
x= -36, y = 15
Thus, the side BC and CA intersect at the point C (-36, 15)
Now,
BC = a = 39
CA = b = 56
AB = c = 25
Similarly, (x1, y1) = A(20, 15)
CA = b = 56
AB = c = 25
Similarly, (x1, y1) = A(20, 15)
(x2, y2) = B(0, 0)
(x3, y3) = C(-36, 15)
(x3, y3) = C(-36, 15)
Therefore, incentre is
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Community Answer
For the triangle whose sides are along the lines y = 15, 3x – 4y...
To find the vertices of the triangle, we need to find the intersection points of the three lines y = 15, 3x - 2y = 0, and 3x + y = 0.
First, let's find the intersection point of y = 15 and 3x - 2y = 0.
Substituting y = 15 into 3x - 2y = 0, we have:
3x - 2(15) = 0
3x - 30 = 0
3x = 30
x = 10
So the first vertex is (10, 15).
Next, let's find the intersection point of y = 15 and 3x + y = 0.
Substituting y = 15 into 3x + y = 0, we have:
3x + 15 = 0
3x = -15
x = -5
So the second vertex is (-5, 15).
Finally, let's find the intersection point of 3x - 2y = 0 and 3x + y = 0.
Solving the system of equations, we have:
3x - 2y = 0
3x + y = 0
Adding the two equations together, we get:
6x - y = 0
6x = y
Substituting 6x for y in 3x - 2y = 0, we have:
3x - 2(6x) = 0
3x - 12x = 0
-9x = 0
x = 0
Substituting x = 0 into 3x + y = 0, we have:
3(0) + y = 0
y = 0
So the third vertex is (0, 0).
Therefore, the vertices of the triangle are (10, 15), (-5, 15), and (0, 0).
First, let's find the intersection point of y = 15 and 3x - 2y = 0.
Substituting y = 15 into 3x - 2y = 0, we have:
3x - 2(15) = 0
3x - 30 = 0
3x = 30
x = 10
So the first vertex is (10, 15).
Next, let's find the intersection point of y = 15 and 3x + y = 0.
Substituting y = 15 into 3x + y = 0, we have:
3x + 15 = 0
3x = -15
x = -5
So the second vertex is (-5, 15).
Finally, let's find the intersection point of 3x - 2y = 0 and 3x + y = 0.
Solving the system of equations, we have:
3x - 2y = 0
3x + y = 0
Adding the two equations together, we get:
6x - y = 0
6x = y
Substituting 6x for y in 3x - 2y = 0, we have:
3x - 2(6x) = 0
3x - 12x = 0
-9x = 0
x = 0
Substituting x = 0 into 3x + y = 0, we have:
3(0) + y = 0
y = 0
So the third vertex is (0, 0).
Therefore, the vertices of the triangle are (10, 15), (-5, 15), and (0, 0).
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For the triangle whose sides are along the lines y = 15, 3x – 4y = 0, 5x + 12y = 0, the incentre is :a)(1, 8)b)(8, 1)c)(-1, 8)d)None of theseCorrect answer is option 'C'. Can you explain this answer? for Class 10 2026 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about For the triangle whose sides are along the lines y = 15, 3x – 4y = 0, 5x + 12y = 0, the incentre is :a)(1, 8)b)(8, 1)c)(-1, 8)d)None of theseCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 10 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the triangle whose sides are along the lines y = 15, 3x – 4y = 0, 5x + 12y = 0, the incentre is :a)(1, 8)b)(8, 1)c)(-1, 8)d)None of theseCorrect answer is option 'C'. Can you explain this answer?.
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