
If the line \[\dfrac{x}{a} + \dfrac{y}{b} = 1\] passes through the points \[(2, - 3)\] and \[(4, - 5)\] then \[(a,b) = \]
A. \[(1,1)\]
B. \[( - 1,1)\]
C. \[(1, - 1)\]
D. \[( - 1, - 1)\]
Answer
161.7k+ views
Hint: In this question, we have to determine the value of \[(a,b)\] by solving the equation \[\dfrac{x}{a} + \dfrac{y}{b} = 1\]by substituting the given points \[(2, - 3)\] and \[(4, - 5)\] in the equation and proceed solving to get the desired solution.
Formula Used: The straight line’s intercept is given by,
\[\dfrac{x}{a} + \dfrac{y}{b} = 1\]
Complete step by step solution: We have been provided in the question that,
An equation \[\dfrac{x}{a} + \dfrac{y}{b} = 1\] passes through the points \[(2, - 3)\] and \[(4, - 5)\]
Because, the given line connects the given places as a result, the given points satisfy the given line equation.
Substitute the values \[x = 2\] and \[y = - 3\] from the given points in the given equation of the line.
On substituting, we get
\[\dfrac{2}{a} - \dfrac{3}{b} = 1\]
On solving the above in equation format, we get
\[ \Rightarrow 2b - 3a = ab\]---- (1)
Now, we have to substitute the values of \[x = 4\] and \[y = - 5\] from the given data in the given equation of the line, we have
\[\dfrac{4}{a} - \dfrac{5}{b} = 1\]
Now, again on solving the above equation, we get
\[ \Rightarrow 4b - 5a = ab\]----- (2)
Now, let us multiply 2 on either side of the equation (1) and we have to subtract from equation (2) , we have
\[4b - 6a - 4b + 5a - 2ab - ab\]
Now we have to cancel all the similar terms, we get
\[ \Rightarrow - a = ab\]
On simplifying the above, we get
\[ \Rightarrow b = - 1\]
Now, let’s multiply 5 on either sides of the equation (1) and we have to multiply 3 on either side of the equation (2) and then we have to subtract, we get
\[10b - 15a - 12b + 15a - 5ab - 3ab\]
Now we have to cancel all the similar terms, we get
\[ \Rightarrow - 2b = 2ab\]
Now, on simplifying the above, we get
\[ \Rightarrow a = - 1\]
Therefore, if the line \[\dfrac{x}{a} + \dfrac{y}{b} = 1\] passes through the points \[(2, - 3)\] and \[(4, - 5)\] then the value of \[(a,b)\] is \[\left( { - 1, - 1} \right)\]
Option ‘D’ is correct
Note: Students must use extreme caution while inserting the values of the x and y-intercepts in the equation of the line while simultaneously satisfying the required requirements. Furthermore, when a point is on a line, it satisfies the line's equation.
Formula Used: The straight line’s intercept is given by,
\[\dfrac{x}{a} + \dfrac{y}{b} = 1\]
Complete step by step solution: We have been provided in the question that,
An equation \[\dfrac{x}{a} + \dfrac{y}{b} = 1\] passes through the points \[(2, - 3)\] and \[(4, - 5)\]
Because, the given line connects the given places as a result, the given points satisfy the given line equation.
Substitute the values \[x = 2\] and \[y = - 3\] from the given points in the given equation of the line.
On substituting, we get
\[\dfrac{2}{a} - \dfrac{3}{b} = 1\]
On solving the above in equation format, we get
\[ \Rightarrow 2b - 3a = ab\]---- (1)
Now, we have to substitute the values of \[x = 4\] and \[y = - 5\] from the given data in the given equation of the line, we have
\[\dfrac{4}{a} - \dfrac{5}{b} = 1\]
Now, again on solving the above equation, we get
\[ \Rightarrow 4b - 5a = ab\]----- (2)
Now, let us multiply 2 on either side of the equation (1) and we have to subtract from equation (2) , we have
\[4b - 6a - 4b + 5a - 2ab - ab\]
Now we have to cancel all the similar terms, we get
\[ \Rightarrow - a = ab\]
On simplifying the above, we get
\[ \Rightarrow b = - 1\]
Now, let’s multiply 5 on either sides of the equation (1) and we have to multiply 3 on either side of the equation (2) and then we have to subtract, we get
\[10b - 15a - 12b + 15a - 5ab - 3ab\]
Now we have to cancel all the similar terms, we get
\[ \Rightarrow - 2b = 2ab\]
Now, on simplifying the above, we get
\[ \Rightarrow a = - 1\]
Therefore, if the line \[\dfrac{x}{a} + \dfrac{y}{b} = 1\] passes through the points \[(2, - 3)\] and \[(4, - 5)\] then the value of \[(a,b)\] is \[\left( { - 1, - 1} \right)\]
Option ‘D’ is correct
Note: Students must use extreme caution while inserting the values of the x and y-intercepts in the equation of the line while simultaneously satisfying the required requirements. Furthermore, when a point is on a line, it satisfies the line's equation.
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