Test: Shifting Of Origin - JEE MCQ

The origin lies on the intersection of X-axis and Y-axis. So, it lies on both the axes.

 Equation : 3x + 4y = 15
Points : (1,-3)
3(x-1) + 4(y-(-3)) = 15
3(x-1) + 4(y+3) = 15
3x - 3 + 4y + 12 = 15
3x + 4y = 6

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 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.

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.

The three points A, B and C are collinear if and only if area of ΔABC = 0.

Midpoint = ((x1 + x2) / 2 , (y1 + y2) / 2) where (x1, y1) and (x2, y2) are the coordinates of the endpoints. Finding the Coordinates of the Centroid: Once we have the midpoints of each side, we can find the equation of the three medians and then find their intersection point. The intersection point is the centroid of the triangle. Let's find the midpoint of each side first: Midpoint of AB = ((-1 + 5) / 2 , (-3 - 6) / 2) = (2, -4.5) Midpoint of AC = ((-1 + 2) / 2 , (-3 + 3) / 2) = (0.5, 0) Midpoint of BC = ((5 + 2) / 2 , (-6 + 3) / 2) = (3.5, -1.5) Now we can find the equation of the medians. The equation of a line that passes through two points (x1, y1) and (x2, y2) is: (y - y1) / (y2 - y1) = (x - x1) / (x2 - x1) Let's find the equation of the median that passes through A: Midpoint of BC = (3.5, -1.5) Slope of BC = (-6 - 3) / (5 - 2) = -3 Equation of BC: y + 6 = -3(x - 5) => y = -3x + 21 Slope of median from A to BC = (3 - (-3)) / (2 - 5) = 2/3 Equation of median from A to BC: y + 3 = (2/3)(x + 1) => y = (2/3)x + (5/3) Similarly, we can find the equations of the medians that pass through B and C: Equation of median from B to AC: y + 6 = (1/3)(x - 5) => y = (1/3)x + 11/3 Equation of median from C to AB: y - 3 = (-2/3)(x - 2) => y = (-2/3)x + 5/3 Now we need to find the intersection point of these three medians. We can solve the system of equations: y = (2/3)x + (5/3) y = (1/3)x + 11/3 y = (-2/3)x + 5/3 Solving these equations, we get x = -1 and y =4

 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)