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# If the midpoint of the line joining (3, 4) and (k, 7) is (x, y) and also passes through 2x+2y+1=0, then find the value of ‘k’.

Last updated date: 23rd May 2024
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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’.

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.