Question

If the distance between two points
A(4, 2), and B(1,y) is 5 units then find
the volue
of y​

Answer 1

Question :

If the distance between two points  A(4, 2) and B(1, y) is 5 units, then find  the values  of y​.

Solution:

We are given two points  P(a, b)  and  Q(c, d).  If we are asked to find the distance between the points P and Q, we may use the formula given below.

$\sqrt{(c-a)^2+(d-b)^2}$

Given that AB = 5 units.

$\Longrightarrow\ \sqrt{(1-4)^2+(y-2)^2}=5\\ \\ \\ \Longrightarrow\ (1-4)^2+(y-2)^2=25\\ \\ \\ \Longrightarrow\ (-3)^2+(y^2-4y+4)=25\\ \\ \\ \Longrightarrow\ 9+y^2-4y+4=25\\ \\ \\ \Longrightarrow\ y^2-4y+13-25=0\\ \\ \\ \Longrightarrow\ y^2-4y-12=0\\ \\ \\ \Longrightarrow\ y^2-6y+2y-12=0\\ \\ \\ \Longrightarrow\ y(y-6)+2(y-6)=0\\ \\ \\ \Longrightarrow\ (y+2)(y-6)=0\\ \\ \\ \therefore\ \ \ \ y=-2\ \ \ ; \ \ \ y=6$

Hence the two possible values for 'y' are  -2  and  6.  

Answer 2

Step-by-step explanation:

d=√{(x2-x1)^2 +(y2-y1)^2}

5 = √ (1-4)^2 + (y-2)^2

5 = √(-3)^2 + (y-2)^2

5 = √{9 + (y-2)^2}

Now squaring both the sides we get

25 = 9+(y-2)^2

25-9 = y^2 +4 -4y

16-4 = y^2-4y

12=y^2-4y

or y^2-4y-12=0

solving quadratic equation

(y+2)(y-6)=0

y= 6, -2