Question

Hi there Brainlians

Please solve the below question and please Can you tell a shortcut method to solve a question like below??

Question ) If the point (-3,a) is the image of the point (1,a+4) in the point (b,1). Find the values of a and b.
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Answer 1

ANSWER:

• The value of a = - 1

• The value of b = - 1

GIVEN:

• The point (- 3, a) is the image of the point (1, a + 4)

• Another point is (b, 1).

TO FIND:

• The values of a and b.

EXPLANATION:

The point (- 3, a) is the image of the point (1, a + 4)

So the point (b, 1) lies between image and the object.

Hence (b, 1) is the midpoint of image and object.

$\boxed{ \bold{MIDPOINT=\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \ \ }}$

$\sf Midpoint = (b, 1)$

$\sf(x_1,\ y_1) = (1, a + 4)$

$\sf(x_2,\ y_2) = ( - 3, a)$

$\sf(b,1)=\dfrac{1+(-3)}{2}, \dfrac{a+4+a}{2}$

$\sf(b,1)=\dfrac{ - 2}{2}, \dfrac{2a + 4}{2}$

$\sf(b,1)= (- 1, a + 2)$

Equate x and y coordinates

b = - 1

1 = a + 2

a = - 1

Hence the value of a = - 1 and b = - 1.

VERFICATION:

Image = (- 3, - 1)

Object = (1, - 1 + 4) = ( 1, 3)

Midpoint = (- 1, 1)

$\boxed{ \bold{MIDPOINT=\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \ \ }}$

$\sf(x_1,\ y_1) = (1, 3)$

$\sf(x_2,\ y_2) = ( - 3, -1)$

$\sf(- 1 , 1) = \dfrac{1-3}{2}, \dfrac{3-1}{2}$

$\sf(- 1 , 1) = \dfrac{-2}{2}, \dfrac{2}{2}$

$\sf(- 1 , 1) = (-1, 1)$

HENCE VERIFIED.

Note : Refer attachment for diagram.

Answer 2

Answer:

a = -1 and b = -1

Step-by-step explanation:

If it was a triangle then we have to find the centroid (mean point of all the positions). But in question no figure is given, only two points are given

Method to find the mean is same. Sum of data with respect to total one.

Assume that point A is (-3, a) and B is (1, a+4) in the point (b,1). Assume that (b,1) as C pint.

(x, y) = (x1 + x2)/2, (y1 + y2)/2

→ b = (-3 + 1)/2

→ 2b = -2

→ b = -1

Similarly,

→ 1 = (a + a + 4)/2

→ 2 = 2a + 4

→ 2 = 2(a + 2)

→ 1 = a + 2

→ a = -1

Hence, the value of a and b is -1.