Question

What is the equation of the line that is perpendicular to the line and passes through the point (15,-5)?

Answer 1

It looks like the equation of the given line was omitted. Please tell me that line.

Meanwhile here is how to find the perpendicular line through (15, −5):

1. If the given line is y = m x + b (m finite), the perpendicular slope is m⊥ = −1/m. Then use point–slope:
y + 5 = (−1/m)(x − 15).

2. If the given line is in standard form ax + by + c = 0 (with a and b not both zero), its slope is −a/b, so the perpendicular slope is b/a. The perpendicular line through (x0,y0) = (15,−5) is
y + 5 = (b/a)(x − 15),
or in standard form: b x − a y + (a·(−5) − b·15) = 0, i.e.
b x − a y − (5a + 15b) = 0.

3. Special cases:
- If the given line is vertical (x = k), the perpendicular line is horizontal: y = −5.
- If the given line is horizontal (y = k), the perpendicular line is vertical: x = 15.

Give me the missing line and I’ll compute the exact equation.