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What is the y-intercept of the equation of the line that is perpendicular to the line y = 3/5x + 10 and passes through the point (15, –5)? The equation of the line in slope-intercept form is y =-5/3 x + ____

Respuesta :

texaschic101
y = 3/5x + 10...slope here is 3/5. A perpendicular line will have a negative reciprocal slope. All that means is flip the slope and change the sign. So we need a slope of -5/3

y = mx + b
slope(m) = -5/3
(15,-5)...x = 15 and y = -5
now we sub for b, the y int
-5 = -5/3(15) + b
-5 = -25 + b
-5 + 25 = b
20 = b

so ur perpendicular equation is : y = -5/3x + 20

[tex]y-[/tex] intercept is [tex]20[/tex] and an equation in the slope- intercept form is [tex]y=\frac{-5}{3}x+20 [/tex]

Slope intercept form of an equation having slope [tex]m[/tex] and [tex]y-[/tex]intercept equal to [tex]b[/tex] is given by [tex]\boldsymbol{y=mx+b}[/tex]

Given equation is [tex]y=\frac{3}{5}x+10 [/tex]

Slope is [tex]\frac{3}{5} [/tex]

The slopes of two perpendicular lines are negative reciprocals of each other.

So, slope of the required line is [tex]\frac{-5}{3} [/tex]

Equation of a line having slope [tex]m'[/tex] and passing through point [tex](x_1,y_1)[/tex] is given by [tex]y-y_1=m(x-x_1)[/tex]

[tex]y+5=\frac{-5}{3}(x-15)[/tex]

[tex] 3y+15=-5x+75[/tex]

[tex]3y=-5x+60[/tex]

[tex]y=\frac{-5}{3}x+20 [/tex]

So, [tex]y-[/tex] intercept is [tex]20[/tex]

For more information:

https://brainly.com/question/3334417?referrer=searchResults

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