Question

What is the equation of the line that is perpendicular to the line y=*x+10 and passes through the point (15,-5)?

Answer 1

y= -x+10

Explanation:

The general equation of a line is

y=mx+b where m is the slope and b is the y-intercept.

The given equation is

y= x+10, where m =1 is the slope and b=10 is the y-intercept

To know: Perpendicular lines have their slope negative reciprocal

y = - x+b is the line ⊥to the given line that has the slope m= -1

To find the y-intercept use the given point (15, -5)

-5 = -15 +b

-5+15 = b

10 = b

So the equation of the line that is perpendicular to the line y= x+10 and passes through the point (15,-5) is y= -x +10

Answer 2

Slope of the line y = x + 10 is 1. So, any straight line perpendicular to this must be having slope of (-1).

Let, the line be y = - x + b and it passes through (15, -5).

So, 15 = 5 + b

b = 10.

So, equation of that line: y = -x + 10

Or, x + y = 10.

Y-intercept of the line = 10 units.