Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

Asked by Pragya Singh | 11 months ago |  77

1 Answer

Solution :-

Points are (2, 5) and (-3, 6).

We know that slope, m = \( \dfrac{y_2-y_1}{x_2-x_1}\)

\( \dfrac{(6 –5)}{(-3-2)}\)

= \( \dfrac{1}{-5}\)

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Then, m = (\( -\dfrac{1}{m}\))

\( \dfrac{-1}{\dfrac{-1}{5}}\)

= 5

We know that the point (x, y) lies on the line with slope m through the fixed point (x0, y0), if and only if, its coordinates satisfy the equation y – y0 = m (x – x0)

Then, y – 5 = 5(x – (-3))

y – 5 = 5x + 15

5x + 15 – y + 5 = 0

5x – y + 20 = 0

The equation of the line is 5x – y + 20 = 0

Answered by Abhisek | 11 months ago

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