Answer

Verified

456k+ views

Hint: First find the midpoint between (3,4) and (k,7) using formula,

\[x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2},y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\] where (x, y) is the midpoint of points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] then put it in the equation 2x +2y +1 = 0 to get the value of ‘K’.

Complete step-by-step answer:

At first we will find the midpoint using the formula

\[x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2},y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\]

Where (x, y) is the midpoint of points \[\left( {{x}_{1}},{{y}_{1}} \right)\]and \[\left( {{x}_{2}},{{y}_{2}} \right)\].

So, if the points are (k,7) and (3,4) then its midpoint will be

\[\left( \dfrac{k+3}{2},\dfrac{7+4}{2} \right)=\left( \dfrac{k+3}{2},\dfrac{11}{2} \right)\]

Now we were given that (x,y) were the mid points of (3,4) and (k,7) then we can say that,

\[\left( \dfrac{k+3}{2},\dfrac{11}{2} \right)=\left( x,y \right)\]

In the question it is given that (x, y) passes through line 2x + 2y + 1 = 0.

So, substituting \[x=\dfrac{k+3}{2},y=\dfrac{11}{2}\] in equation 2x + 2y + 1 = 0, we get,

\[2\left( \dfrac{k+3}{2} \right)+2\left( \dfrac{11}{2} \right)+1=0\]

On further calculations we get,

$\Rightarrow$ k + 3 + 11 + 1 = 0

$\Rightarrow$ k + 15 = 0

$\Rightarrow$ k = -15

Therefore, the required value of ‘k’ is ‘-15’.

Note: Students after finding out midpoint they generally get confused about how to find ‘k’. If a line is passing through another line, then the intersection point is the same. So, read the question thoroughly before solving it and also be careful about calculation errors. Another approach is finding the equation of line passing through the points (3, 4) and (k, 7), then finding the intersection point of this line with 2x+2y+1=0.

\[x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2},y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\] where (x, y) is the midpoint of points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] then put it in the equation 2x +2y +1 = 0 to get the value of ‘K’.

Complete step-by-step answer:

At first we will find the midpoint using the formula

\[x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2},y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\]

Where (x, y) is the midpoint of points \[\left( {{x}_{1}},{{y}_{1}} \right)\]and \[\left( {{x}_{2}},{{y}_{2}} \right)\].

So, if the points are (k,7) and (3,4) then its midpoint will be

\[\left( \dfrac{k+3}{2},\dfrac{7+4}{2} \right)=\left( \dfrac{k+3}{2},\dfrac{11}{2} \right)\]

Now we were given that (x,y) were the mid points of (3,4) and (k,7) then we can say that,

\[\left( \dfrac{k+3}{2},\dfrac{11}{2} \right)=\left( x,y \right)\]

In the question it is given that (x, y) passes through line 2x + 2y + 1 = 0.

So, substituting \[x=\dfrac{k+3}{2},y=\dfrac{11}{2}\] in equation 2x + 2y + 1 = 0, we get,

\[2\left( \dfrac{k+3}{2} \right)+2\left( \dfrac{11}{2} \right)+1=0\]

On further calculations we get,

$\Rightarrow$ k + 3 + 11 + 1 = 0

$\Rightarrow$ k + 15 = 0

$\Rightarrow$ k = -15

Therefore, the required value of ‘k’ is ‘-15’.

Note: Students after finding out midpoint they generally get confused about how to find ‘k’. If a line is passing through another line, then the intersection point is the same. So, read the question thoroughly before solving it and also be careful about calculation errors. Another approach is finding the equation of line passing through the points (3, 4) and (k, 7), then finding the intersection point of this line with 2x+2y+1=0.

Recently Updated Pages

Find the circumference of the circle having radius class 8 maths CBSE

Is the equation 6x2 + 3y 0an example of direct var class 8 maths CBSE

Why are there 2 pi radians in a circle class 8 maths CBSE

What is the remainder of 329 divided by 4 class 8 maths CBSE

The price of a pound of peppers is 399dollar What is class 8 maths CBSE

What is the greatest common divisor GCD of 78 and class 8 maths CBSE

Trending doubts

Write a letter to the principal requesting him to grant class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

One cusec is equal to how many liters class 8 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Give 10 examples of Material nouns Abstract nouns Common class 10 english CBSE

Write an application to the principal requesting five class 10 english CBSE

10 examples of evaporation in daily life with explanations

A Short Paragraph on our Country India