(a) Find the equation of the normal to the curve y = (x\(^2\) - x + 1)(x - 2) at the point where the curve cuts the X - axis.
(b) The coordinates of the pints P, Q and R are (-1, 2), (5, 1) and (3, -4) respectively. Find the equation of the line joining Q and the midpoint of \(\overline{PR}\).
(b) The coordinates of the pints P, Q and R are (-1, 2), (5, 1) and (3, -4) respectively. Find the equation of the line joining Q and the midpoint of \(\overline{PR}\).
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Correct Answer: Option
Explanation:
(a) From y = (x\(^2\) - x + 1)(x - 2) - x + 1)(x - 2)
= x\(^3\)- 3x\(^2\) + 3x - 2
dy/dx = 3x\(^2\) - 6x + 3
=x\(^2\) - 2x + 1
(x - 1) = 0 twice ; x = 1
y = (1\(^2\) - 1 + 1) (1-2) ; y = -1
Gradient m\(_1\) of tangent = y/x = -1/1 = -1
Using m\(_1\)m\(_2\) = -1
Gradient, m\(_2\) of normal =1
Using y-y\(_1\) = m\(_2\)(x - x\(_1\))
y + 1 = x - 1; x - y - 2 = 0
x\(_1\),y\(_1\) = ( -1, 2), , x\(_2\),y\(_2\), =(3, -4)
Mid point of PR =\(\frac{3-1,-4+2}{2,2}\)
= (1-1)
Using \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{y - y_1}{x - x_1}\)
\(\frac{1-{-1}}{5-1}\) = \(\frac{y-1}{x-5}\);
\(\frac{2}{4}\) = \(\frac{y-1}{x-5}\)
2(y-1) = x - 5; 2y - 2 = x - 5
2y - x = -3
x - 2y -3 =0
(a) From y = (x\(^2\) - x + 1)(x - 2) - x + 1)(x - 2)
= x\(^3\)- 3x\(^2\) + 3x - 2
dy/dx = 3x\(^2\) - 6x + 3
=x\(^2\) - 2x + 1
(x - 1) = 0 twice ; x = 1
y = (1\(^2\) - 1 + 1) (1-2) ; y = -1
Gradient m\(_1\) of tangent = y/x = -1/1 = -1
Using m\(_1\)m\(_2\) = -1
Gradient, m\(_2\) of normal =1
Using y-y\(_1\) = m\(_2\)(x - x\(_1\))
y + 1 = x - 1; x - y - 2 = 0
x\(_1\),y\(_1\) = ( -1, 2), , x\(_2\),y\(_2\), =(3, -4)
Mid point of PR =\(\frac{3-1,-4+2}{2,2}\)
= (1-1)
Using \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{y - y_1}{x - x_1}\)
\(\frac{1-{-1}}{5-1}\) = \(\frac{y-1}{x-5}\);
\(\frac{2}{4}\) = \(\frac{y-1}{x-5}\)
2(y-1) = x - 5; 2y - 2 = x - 5
2y - x = -3
x - 2y -3 =0