Find a point symmetrical to the point М (2, -1) relative to the line x-2y + 3 = 0

| August 6, 2021 | education

We represent the equation of a straight line in the form l1: x-2y + 3 = 0, -2y = -x-3, y = 1 / 2x + 3/2. Its slope is k1 = 1/2. Hence, the slope of the line l2 perpendicular to l1 is k2 = −1 / k1 = -1 / 1/2 = -2. We write the equation of the straight line l2 passing through the point M (2: −1): y = -2x + c, -1 = -2 * 2 + c, c = 3, then y = -2x + 3. Let’s solve the equations of straight lines together, find the point of their intersection: y = -2x + 3, y = 1 / 2x + 3/2, 1 / 2x + 3/2 = -2x + 3, 5 / 2x = 3/2, x = 3/5, y = 1/2 * 3/5 + 3/2, y = 3/10 + 3/2, y = 9/5. Point A (3/5; 9/5) will be the midpoint of the MC segment, if we denote the desired point by C. Find point C, solve the system of equations: 3/5 = (2 + x) / 2, 9/5 = (-1+ y) / 2, 10 + 5x + 6, 54 = -4, x = -4/5, -5 + 5y = 18, 5y = 23, y = 23/5.
Answer: a point symmetrical to the point M (2, -1) relative to the straight line x-2y + 3 = 0 – C (-4/5; 23/5)



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