Please help this is my last question also explain it if you could thanks.

Line segment segment AB has a slope of start fraction four over three end fraction and contains point A(6, −5). What is the y-coordinate of point Q(2, y) if segment QA is perpendicular to line segment segment AB question mark

y = −2

y = −1

y = 2

y = −3

The y-coordinate of point Q is y = -2, hence, point Q's coordinate is (2,-2).

Step-by-step explanation:

The relationship between two perpendicular lines with slopes m₁ and m₂ is given by

m₂ = -(1/m₁)

Let m₁ be the slope of line segment AB and m₂ be the slope of line segment QA

The slope of AB is given in the question as m₁ = 4/3

The slope of QA, m₂ = -(1/(4/3)) (since the two lines are perpendicular)

So, m₂ = -(3/4)

But the slope of any line can be obtained from the coordinates of two known points on that line.

Slope of a line with two points (x₁,y₁) and (x₂,y₂) on it, is given as (y₂ - y₁)/(x₂ - x₁)

For line segment QA with two points Q (2,y) and A(6,-5) known,

Slope of QA, m₂ = (-3/4) = (-5 - y)/(6 - 2)

(-3/4) = (-5 - y)/(4)

-3 = -5 - y

y = -5 + 3 = -2

Hope this Helps!!

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation:

y= 2x+7

changed to y= 5x +7

as slope is increased from 2 to 5

therefore it would be less steeper

so, option d) is correct