Question

What is the y-intercept of the equation of the line that is perpendicular to the line v= x -x+ 10 and passes through the

point (15,-5)?
O y=xx-20
O y=-3x+20
Ov=3x-2
Ov=-3x+20

Answer 1

`y=-(5)/(3)x+20`

Here, m=-5/3, and b=y-intercept=20

Here, the y-intercept is: 20

Thus, option (d) is true.

Explanation:

Given the equation

`y=(3)/(5)x+10`

comparing the equation with the slope-intercept form

`y=mx+b`

Here,

• m is the slope

• y is the intercept

so the slope of the line is 3/5.

As we know that the slope of the perpendicular line is basically the negative reciprocal of the slope of the line, so

The slope of the perpendicular line will be: -5/3

Therefore, the point-slope form of the equation of the perpendicular line that goes through (15,-5) is:

`y-y_1=m\left(x-x_1\right)`

`y-\left(-5\right)=(-5)/(3)\left(x-15\right)`

`y+5=(-5)/(3)\left(x-15\right)`

simplifying the equation to convert it into the slope-intercept form

We know that the slope-intercept form of the line equatio is

`y=mx+b`

here 'm' is the slope and 'b' is the y-intercept

`y+5=(-5)/(3)\left(x-15\right)`

subtract 5 from both sides

`y+5-5=(-5)/(3)\left(x-15\right)-5`

`y=-(5)/(3)x+20`

Here, m=-5/3, and b=y-intercept=20

Here, the y-intercept is: 20

Thus, option (d) is true.