Prove that the line through the point \((x_1, y_1)\) and parallel to the line \(Ax + By + C = 0 \) is \(A (x -x_1) + B (y - y_1) = 0.\)

Question

If the image of the point \( P(1, 2, a) \) in the line \[ \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2} \] is \( Q(5, b, c) \), then \( a^2 + b^2 + c^2 \) is equal to

Answer 1

The slope of line \(Ax + By + C = 0\) or \(y =\frac{ -A}{B}x –\frac{ C}{B}\) is \(m = \frac{-A}{B}\)
It is known that parallel lines have the same slope.

∴Slope of the other line \(=m = \frac{-A}{B}\)
The equation of the line passing through point \((x_1, y_1) \) and having a slope \(m = \frac{-A}{B}\) is

\(y – y_1 = m (x – x_1)\)

\(y – y_1= \frac{-A}{B }(x – x_1)\)

\(B (y – y_1) = -A (x – x_1)\)

\(∴ A(x – x_1) + B(y – y_1) = 0\)

Hence, the line through point \((x_1, y_1)\) and parallel to line \(Ax + By + C = 0\) is \(A (x - x_1) + B (y - y_1) = 0\)

Prove that the line through the point \((x_1, y_1)\) and parallel to the line \(Ax + By + C = 0 \) is \(A (x -x_1) + B (y - y_1) = 0.\)

Solution and Explanation

Top Questions on Distance of a Point From a Line

Questions Asked in CBSE Class XI exam