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Question: We have two given points as \( P\left( {0,5} \right) \) , \( Q\left( {15, - 4} \right) \). If the po...

We have two given points as P(0,5)P\left( {0,5} \right) , Q(15,−4)Q\left( {15, - 4} \right). If the point (5,a)\left( {5,a} \right) lies on the line PQPQ then the value of aa is:
A. 44
B. 55
C. 32\dfrac{3}{2}
D. 22

Explanation

Solution

Hint : Here, we are given two points on the same line. First, we need to find the equation of the line using these two given points which can be obtained after finding the slope of the line using the given points. After that, we will put the point (5,a)\left( {5,a} \right) in the equation of line to find the value of aa .
Formulas used:
When two points (x1,y1)({x_1},{y_1}) and (x2,y2)({x_2},{y_2}) are on the same line, then the slope of the line is given by
m=y2−y1x2−x1m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}
And the equation of line is given by
y−y1=m(x−x1)y - {y_1} = m\left( {x - {x_1}} \right)

Complete step-by-step answer :
We are given that P(0,5)P\left( {0,5} \right) and Q(15,−4)Q\left( {15, - 4} \right) are on the line PQPQ . Therefore, for finding the equation of line PQPQ , first we will find the slope of the line using both the points.
We know that slope of the line is given by m=y2−y1x2−x1m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}
Here, we have (x1,y1)=(0,5)({x_1},{y_1}) = \left( {0,5} \right) and (x2,y2)=(15,−4)({x_2},{y_2}) = \left( {15, - 4} \right)
⇒m=−4−515−0=−915=−35\Rightarrow m = \dfrac{{ - 4 - 5}}{{15 - 0}} = \dfrac{{ - 9}}{{15}} = - \dfrac{3}{5}
Now, we will determine the equation of line PQPQ by using this slope value.
We know that equation of line is y−y1=m(x−x1)y - {y_1} = m\left( {x - {x_1}} \right)
Here, we have (x1,y1)=(0,5)({x_1},{y_1}) = \left( {0,5} \right) and m=−35m = - \dfrac{3}{5}
⇒y−5=−35(x−0) ⇒5y−25=−3x ⇒3x+5y−25=0   \Rightarrow y - 5 = - \dfrac{3}{5}\left( {x - 0} \right) \\\ \Rightarrow 5y - 25 = - 3x \\\ \Rightarrow 3x + 5y - 25 = 0 \;
Thus, the equation of line PQPQ is 3x+5y−25=03x + 5y - 25 = 0
Now, we will put (x,y)=(5,a)\left( {x,y} \right) = \left( {5,a} \right) in this equation.
3x+5y−25=0 ⇒3(5)+5(a)−25=0 ⇒15+5a−25=0 ⇒5a−10=0 ⇒5a=10 ⇒a=105 ⇒a=2   3x + 5y - 25 = 0 \\\ \Rightarrow 3\left( 5 \right) + 5\left( a \right) - 25 = 0 \\\ \Rightarrow 15 + 5a - 25 = 0 \\\ \Rightarrow 5a - 10 = 0 \\\ \Rightarrow 5a = 10 \\\ \Rightarrow a = \dfrac{{10}}{5} \\\ \Rightarrow a = 2 \;
Thus, the value of aa is 2.
So, the correct answer is “a=2”.

Note : In this problem, we have determined the equation of line using the formula y−y1=m(x−x1)y - {y_1} = m\left( {x - {x_1}} \right) . However, we can also use the standard line equation y=mx+cy = mx + c and put the values of both given points in it respectively. In this way, we can obtain two equations by solving which we can determine the values of mm and cc . Thus after getting the equation of line, we can put the point (5,a)\left( {5,a} \right) in that equation of line to find the value of aa .