The problem asks us to find the coordinates of point B, given the coordinates of point A (2, 4) and the midpoint D (-1, 3) of the line segment AB.
Let the coordinates of A be \((x_1, y_1) = (2, 4)\).
Let the coordinates of B be \((x_2, y_2)\).
Let the coordinates of the midpoint D be \((x_m, y_m) = (-1, 3)\).
The midpoint formula states that:
\((x_m, y_m) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
Equate x-coordinates:
Use the x-coordinate part of the midpoint formula:
\(x_m = \frac{x_1 + x_2}{2}\)
Substitute the known values:
\(-1 = \frac{2 + x_2}{2}\)
Multiply both sides by 2:
\(-1 \times 2 = 2 + x_2\)
\(-2 = 2 + x_2\)
Solve for \(x_2\):
\(x_2 = -2 - 2\)
\(x_2 = -4\)
Equate y-coordinates:
Use the y-coordinate part of the midpoint formula:
\(y_m = \frac{y_1 + y_2}{2}\)
Substitute the known values:
\(3 = \frac{4 + y_2}{2}\)
Multiply both sides by 2:
\(3 \times 2 = 4 + y_2\)
\(6 = 4 + y_2\)
Solve for \(y_2\):
\(y_2 = 6 - 4\)
\(y_2 = 2\)
Conclusion:
The calculated coordinates for point B are \((-4, 2)\).
Comparing the calculated coordinates \((-4, 2)\) with the given options, Option 3 matches our result.
The graphs of the linear equations 4x - 2y = 10 and 4x + ky = 2 intersect at a point (a, 4). The value of k is equal to:
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The area (in sq. units) of the triangle formed by the graphs of 8x + 3y = 24, 2x + 8 = y and the x-axis is:
In which quadrant both abscissa and ordinate are negative?
Find the slope of the line joining the points (3, -4) and (5, 2).