The origin lies on
The origin lies on the intersection of X-axis and Y-axis. So, it lies on both the axes.
What will be the new equation of the straight line 3x + 4y = 15, if the origin gets shifted to (1,-3) ?
Equation : 3x + 4y = 15
Points : (1,-3)
3(x-1) + 4(y+3) = 15
3x - 3 + 4y + 12 = 15
3x + 4y = 6
What will be the new equation of the straight line 5x + 8y = 10, if the origin gets shifted to (2,-3) ?
Equation : 5x + 8y = 10
Points (2, -3)
(x-2, y+3)
⇒ 5(x-2) + 8(y+3) = 10
= 5x - 10 + 8y + 24 = 10
⇒ 5x + 8 = - 4
The point where all the angle bisectors of a triangle meet is
The angle bisectors of the angles of a triangle are concurrent (they intersect in one common point). The point of concurrency of the angle bisectors is called the incenter of the triangle.
What will be the new equation of the straight line 5x + 8y = 10, if the origin gets shifted to (2,-3) ?
What will be the value of ‘p” if the equation of the straight line 3x + 5y = 10 gets changed to 3x + 5y = p after shifting the origin at (2,2) ?
3x + 5y = 10, at origin
But now, it’s (2,2),
3(x-2) + 5(y-2) = 10
Hence, 3x - 6 + 5y - 10 = 10
3x + 5y = 26
So, 3x + 5y = p
=> p = 26.
If area of ΔABC = 0 ,three points A,B,C are
The three points A, B and C are collinear if and only if area of ΔABC = 0.
The distance between the pair of points (7,8) and (4,2) ,if origin is shifted to (1,-2) ,would be
The coordinates of centroid of triangle whose vertices are A(-1,-3), B(5,-6) and C(2,3) and origin gets shifted to (1,2)
New coordinates of the point (7,-1) would be, if the origin is shifted to the point (1,2) by translation of the axis.
Let the new origin be (h, k) = (1, 2) and (x, y) = (7, -1) be the given point
Therefore new co-ordinates (X, Y) .
x = X + h and y = Y + k
i.e. 7 = X + 1 and -1 = Y + 2
This gives, X = 6 and Y = -3.
Thus the new co-ordinates are (6, -3)
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