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Question: Let A = \[{\text{ }}\left( {3, - 4{\text{ }}} \right)\], B = \[\left( {1,2} \right)\]. Let P = \[\le...

Let A =  (3,−4 ){\text{ }}\left( {3, - 4{\text{ }}} \right), B = (1,2)\left( {1,2} \right). Let P = (2k−1 , 2k+1 )\left( {2k - 1{\text{ }},{\text{ }}2k + 1{\text{ }}} \right)be a variable point such that PA + PB is the minimum . Then k is:

Explanation

Solution

Hint : In order to solve the given question, we must have the knowledge of basics of straight line where we study about slope also which comes under coordinate geometry. The Slope of a line determines a number that measures its steepness or mathematically we can say that it is the change in y for a unit change in x along the line . When any two points on the line is given , the slope of a line can be calculated by the formula m=y−y1x−x1m = \dfrac{{y - {y_1}}}{{x - {x_1}}}

Complete step by step solution:
In the question given , we need to find the minimum value of PA + PB when we are given the points of A, B and P respectively as –
Let A =  (3,−4 ){\text{ }}\left( {3, - 4{\text{ }}} \right)
B = (1,2)\left( {1,2} \right)
P = (2k−1 , 2k+1 )\left( {2k - 1{\text{ }},{\text{ }}2k + 1{\text{ }}} \right)
We can get the minimum value of PA + PB iff point P lies on the line AB or the joining points A and B .
So, we will equate the slopes and get our answer as follows by using the formula of slope = m=y−y1x−x1m = \dfrac{{y - {y_1}}}{{x - {x_1}}}

2k+1−22k−1−1=−4−23−1 2k+12k−2=−62 2k−12k−2=−3 2k−1=−3(2k−2) 2k−=−6k+6   \dfrac{{2k + 1 - 2}}{{2k - 1 - 1}} = \dfrac{{ - 4 - 2}}{{3 - 1}} \\\ \dfrac{{2k + 1}}{{2k - 2}} = \dfrac{{ - 6}}{2} \\\ \dfrac{{2k - 1}}{{2k - 2}} = - 3 \\\ 2k - 1 = - 3(2k - 2) \\\ 2k - = - 6k + 6 \;

To simplify we will do calculations adding and subtracting both the sides of the equation as per our requirement .

6k+2k=6+1 8k=7 k=78   6k + 2k = 6 + 1 \\\ 8k = 7 \\\ k = \dfrac{7}{8} \;

So, the correct answer is “ k=78k = \dfrac{7}{8} ”.

Note : In equivalent equations which have identical solutions we can perform multiplication or division by the same non-zero number both L.H.S. and R.H.S. of an equation .
In an equivalent equation which has an identical solution we can raise the same odd power to both L.H.S. and R.H.S. of an equation .
Always try to understand the mathematical statement carefully and keep things distinct .
Choose the options wisely , it's better to break the question and then solve part by part .
Cross check the answer and always keep the final answer simplified .