Question

the equation of a curve is y=x^2-4x+7 and the equation of a line is y+3x=9. the curve and the line intersect at the points A and B.
the midpoint of AB is M. show the coordinates of M are (1/2,7 and 1/2)

5ad0522463b3c803a0be05d7a45be24b.docx

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• The equation of a curve is y=x^2 - 4x + 7 and the equation of a line is y + 3x = 9.
• The curve and the line intersect at the points A and B.
• The midpoint of AB is M.
• Show the coordinates of M are (1/2, 7 and 1/2)

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$\large\underline{\bold{❥︎Step :- 1 }}$

☆ To calculate the point of intersection of curve and line.

$\tt \:  ⟼ \: Let \: y = {x}^{2} - 4x + 7 \: - - - (1)$

$\tt \:  ⟼ \: and \: Let \: y \: = 9 - 3x \: - - - (2)$

☆ On substituting, equation (2) in (1), we get

$\tt \:  ⟼ \: {x}^{2} - 4x + 7 = 9 - 3x$

$\tt \:  ⟼ \: {x}^{2} - x - 2 = 0$

$\tt \:  ⟼ \: {x}^{2} - 2x + x - 2 = 0$

$\tt \:  ⟼ \: x(x - 2) + 1(x - 2) = 0$

$\tt \:  ⟼ \: (x - 2)(x + 1) = 0$

$\tt\implies \: \: x = - 1 \: or \: x = 2$

So, point of intersection are represented in table

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y = 9 - 3x\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf - 1 & \sf 12 \\ \\ \sf 2 & \sf 3 \\ \\ \sf & \sf \end{array}} \\ \end{gathered}$

$\large\underline{\bold{❥︎Step :- 2 }}$

$\tt \:  Let \: the \: points \: be \: A(-1, 12) \: and \: B(2, 3)$

$\tt \:  So, \: midpoint \: M \: of \: AB = (\dfrac{ - 1 + 2}{2} ,\dfrac{3 + 12}{2} )$

$\tt \:  ⟼ \: midpoint \: M \: of \: AB = (\dfrac{1}{2} , \dfrac{15}{2} )$

$\tt\implies \:So, coordinates \: of \: M = ( \dfrac{1}{2} , 7\dfrac{1}{2} )$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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