Question

If the mid-point of the line segment joining the points A (3,4) and B (k,6) is P (x, y) and x + y - 10 = 0, find the value of k.

$\sf{Spam = Report \checkmark}$

Answer 1

Q) If the mid-point of the line segment joining the points A (3,4) and B (k,6) is P (x, y) and x + y - 10 = 0, find the value of k.

ANSWER:-

Mid point of AB is P , where A(3,4) , B(k,6) and P(x,y). therefore:-

x = (3+k)/2. and y = (4+6)/2=5.

Putting x = (3+k)/2 and y = 5. in x + y -10 = 0.

or, (3+k)/2 + 5 -10 = 0.

or, (3+k)/2 = 5.

or, 3 + k = 10.

or, k = 10 - 3 = 7. Answer.

Answer 2

Step-by-step explanation:

$\bigstar$ $\large\pmb{\mathfrak{To \ Find:-}}$

$\implies$ Value of k.

$\underline\mathtt{\red{Let's \ do \ it!!}}$

$\bigstar$ $\large\pmb{\mathfrak{Solution:-}}$

$\bigstar$ Given; P(x,y) is the mid point of AB

• x + y - 10 = 0 or x + y = 10

$\implies$ P = $\frac{A + B}{2}$

Putting the values we have;

$\implies$ P = ($\frac{3 + k}{2}$ , $\frac{6 + 4}{2}$)

$\implies$ P = ($\frac{3 + k}{2}$ , $\frac{10}{2}$)

$\implies$ P = ($\frac{3 + k}{2} , 5)$

$\implies$ x = $\frac{3 + k}{2}$ and y = 5

Now, given that x + y = 10

Putting the values of x and y we have;

$\frac{3 + k}{2}$ + 5 = 10

3 + k + 10 = 20 [Taking the LCM and moving to the RHS]

k + 13 = 20

k = 7

$\bigstar$ $\large\pmb{\mathfrak{More\ to \ know :-}}$

• The distance between points P($x_1$ , $y_1$) and Q($x_2$ , $y_2$) is given by

____________________________

PQ = √ (x_2 - x_1)^2 + (y_2 - y_1)^2