HELP ME PLZZZZZ

Question

How to write Equation of a line that is perpendicular to the line y=3x-5 and passes through the point (5,10)

Answer 1

The current equation is of the form y = mx + b where m is the slope. For this case, m = 3.

The perpendicular slope, m⊥ = -1/m = -1/3.

Therefore, the new equation can be written as follows: y⊥ = -1/3x + c, where c is a new constant to be determined.

Think about it this way: Every time x increases by 1 unit, y decreases by (1/3) units. This means x must be increased by 3 units for y to be decreased by 1 unit. Therefore, if x is increased 5 units to the right, y must decrease 5*(1/3) units or 5/3 units. To get back to the y intercept (where x = 0), x must decrease by 5 units to the left or y must increase by (5/3) units. **(-5/3*(5) + -5/3(-5)) = 0)

So, if we allow the perpendicular line to cross the ordered pair

(5,10: x = 5, y = 10), then it must cross the following ordered pair as well: (5-5, 10 + 5/3: x = 0, y = 35/3)

This means the y-intercept, c = 35/3 --> y⊥ = (-1/3)x + 35/3 or 3*y⊥ + x = 35 **(multiply both sides by 3 then add x to both sides)

Solving algebraically:

If y⊥ = (-1/3)x + c crossed through the ordered pair (5,10), then at x = 5, y⊥ must equal 10.

So, 10 = (-1/3)(5) + c **(Note: I inputted the ordered pair inputs into the equation).

10 = -5/3 + c

+5/3. +5/3. **(add 5/3 to both sides)

30/3 + 5/3 = c --> 35/3 = c. **(10 = 30/3)

So, y⊥ = -1/3x + 35/3

Answer 2

y=(-1/3)x + b where b = y intercept and (-1/3) is the negative inverse of 3=slope of the original equation

plug in the point to find b

10=-5/3 + b

b = 35/3

y= (-1/3)x + 35/3

3y =-x + 35

x+3y = 35