Find the coordinate of the point m(0,a) if m is midpoint between A(3,2)  and B(b,-4)

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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m(0,a ) is midpoint between A(3,2) and (B(b, -4)

We know that if m is midppoint of AB , then the coordinates of the mid point m is:

xM = (xA+xB)/2

xy = (yA+yB)/2

Let us subsitute :

For the x coordinate:

0 = (3+b)/2

==> b= -3

Now , for y coordinate:

a = (2-4)/2 = -2/2 = -1

==> a= -1

Then m(0, -1) is midpoint between A(3, 2) and B(-3,-4)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To calculate the coordinates a and b, we'll write the equation of the line that passes AB.

(xB - xA)/(x-xA) = (yB-yA)/(y-yA)

We'll substitute the coordinates for A and B:

(b-3)/(x-3) = (-4-2)/(y-2)

(b-3)/(x-3) = -6/(y-2)

We'll cross multiply and we'll get:

-6x + 18 = by - 2b - 3y + 6

We'll move all terms to the left side:

-6x + 18 - by + 2b + 3y - 6 = 0

We'll combien like terms:

-6x + y(-b + 3) + 12 + 2b = 0

We know that the point (0,a) belongs to the line AB.

-6*0 + a(-b + 3) + 12 + 2b = 0

a(-b + 3) + 12 + 2b = 0

To verify the solutions for a and b, we'll substitute in the expression:

-ab + 3a + 12 + 2b = 0

We know that (0,a) is the midpoint so:

xM = (xA+xB)/2

xM = (b+3)/2

But xM = 0

b + 3 = 0

b = -3

yM = (yA+yB)/2

yM = (-4+2)/2

But yB = a

a = -1

Verify:

-(-1)(-3) + 3(-1) + 12 + 2(-3) = 0

-3 - 3 + 12 - 6 = 0

-12 + 12 = 0

So, the values for a and b are: {-1 ; -3}.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The coordinates of the mid point M of the points  (x1, y1)  and  (x2, y2) is given by the formula:

Mx = (x1+x2)/2

My = (y1+y2)/2.

Therefore the mid point of A (3, 2) and B(b ,-4) is given by:

0 = (Ax+By)/2 = (3+b)/2 ..(1)and a = (Ay+By)/2 = (2-4)/2....(2)

So  from the first of the equation 3+b = 0 .

 b= -3.

From the secod equation we get:

a = (2-4)/2 = -2.

Therefore a = -2 and b = -3.

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