Find the image of the point (1, 2) in the line x - 2y - 7 = 0.

The given line is x - 2y - 7 = 0 .....(i) ⇒ 2y = x - 7 ⇒ y = . The slope of the line (i) = m 1 = Let the point (1, 2) be P. From P draw a perpendicular to the line (i) and produce it to point P' such that P'M = MP, then P' is the image of P in line (i) and line (i) is the right bisector of the segment PP'. Let P' be (a, b). Then slope of PP' = m 2 = . Since, line (i) is perpendicular to PP' so, Also mid-point of PP' is M. Since, (i) is the right bisector of the segment PP', M lies on (i) Multiplying equation (iv) by 2 and subtracting from (iii) we get, Putting value of b in Eq (iii), ⇒ 2a - 6 = 4 ⇒ 2a = 10 ⇒ a = 5. P' = (a, b) = (5, -6). Hence, the coordinates of image are (5, -6).

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Mathematics

Find the image of the point (1, 2) in the line x - 2y - 7 = 0.

Straight Line Eq

Related Questions

The points B(1, 3) and D(6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.

ABCD is a rhombus. The coordinates of A and C are (3, 6) and (-1, 2) respectively. Write down the equation of BD.

If the line x - 4y - 6 = 0 is the perpendicular bisector of the line segment PQ and the coordinates of P are (1, 3), find the coordinates of Q.

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