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Consider the line segment with endpoints R(-4, -7) and T6, -1) and midpoint (1, -4).What improper fraction (fraction greater than one) represents the y-intercept of the line perpendicular to line segment RTthrough Point S?

Consider The Line Segment With Endpoints R4 7 And T6 1 And Midpoint 1 4What Improper Fraction Fraction Greater Than One Represents The Yintercept Of The Line Pe class=

Sagot :

The line segment RT has coordinates at R(-4, -7) and T(6, -1). Find the slope of this line using the formula;

m= change in y-coordinate values / change in x-coordinate values

m= -1 --7 / 6--4

m= -1+7 /10

m=6/10 --------you can simplify to 3/5

The slope of segment RT is m1= 3/5

The line perpendicular segment RT will have a slope m2 where ;

m1*m2 = -1

Find m2 , which is the gradient of the line perpendicular to segment RT

3/5 * m2 = -1

m2= -5/3

This means the gradient of the perpendicula line to the line segment RT is ;

m2 = -5/3

Now using the slope m2 = -5/3 , the midpoint S, (1, -4) , and imaginary point on the line (x,y) then the equation will be;

m= change in y-values / change in x-values

-5/3 = y--4/x-1

-5/3 = y+4 /x-1

-5{x-1} = 3{y+4}

-5x+5 =3y +12 ------collect like terms and write in form of y= mx + c

-5x+5-12=3y

-5x-7 =3y

-5/3 x - 7/3 = y

y= -5/3 x - 7/3

y-intercept is - 7/3

Answer

-7/3

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