Question

Write the equation of the line that is perpendicular to y=5x+7 and passes through the point (10,3)

Answer 1

Y=-1/5x+5

Explanation:

Y=5x+7

Gradient =5

For perpendicular lines, M1×M2= -1

Therefore, 5×M2= -1(divide both sides by 5)

M2= - 1/5

(10,3) (x, y)

Y-3/x-10= - 1/5 (crossmultiply)

5y-15=-x+10

5y=-x+10+15

5y= - x+25 (divide all sides by 5)

Y= - 1/5x+5

Answer 2

Slope-intercept form: y = mx + b

(m is the slope, b is the y-intercept or the y value when x = 0 --> (0, y) or the point where the line crosses through the y-axis)

For lines to be perpendicular, their slopes have to be negative reciprocals of each other. (flip the sign +/- and the fraction(switch the numerator and the denominator))

For example:

Slope = 2 or
`(2)/(1)`

Perpendicular line's slope =
`-(1)/(2)` (flip the sign from + to -, and flip the fraction)

Slope =
`-(4)/(5)`

Perpendicular line's slope =
`(5)/(4)` (flip the sign from - to +, and flip the fraction)

y = 5x + 7 The slope is 5, so the perpendicular line's slope is
`-(1)/(5)`

Now that you know the slope, substitute/plug it into the equation:

y = mx + b

`y=-(1)/(5) x+b` To find b, plug in the point (10, 3) into the equation, then isolate/get the variable "b" by itself

`3=-(1)/(5) (10)+b`

3 = -2 + b Add 2 on both sides to get "b" by itself

3 + 2 = -2 + 2 + b

5 = b

`y=-(1)/(5) x+5`