Equation of line in Straight lines

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Question

The value of a for which the image of the point $(a, a - 1)$ w.r.t. the line mirror $3 x + y = 6 a$ is the point $(a^{2} + 1, a)$ is

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Solution

Image of point P in line AB is Q.
Thus, L is the midpoint of AB.
$∴ L \equiv (\frac{a^{2} + a + 1}{2}, \frac{2 a - 1}{2})$
Since L lies on line AB.
$\Rightarrow 3 a^{2} - 7 a + 2 = 0$
$∴ a = 2, \frac{1}{3}$
Also, $PQ ⊥ AB .$
$\Rightarrow - 3 \times \frac{1}{a^{2} - a + 1} = - 1$
$\Rightarrow a^{2} - a - 2 = 0$
$∴ a = 2, - 1$

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Equation of line in Straight lines

Remember concepts with our Masterclasses.

The value of a for which the image of the point a,a−1 w.r.t. the line mirror 3x+y=6a is the point a2+1,a is

Image of point P in line AB is Q.Thus, L is the midpoint of AB.∴L≡a2+a+12,2a−12Since L lies on line AB.⇒ 3a2−7a+2=0∴ a=2,13Also, PQ⊥AB.⇒ −3×1a2−a+1=−1⇒ a2−a−2=0∴ a=2,−1

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The value of a for which the image of the point $(a, a - 1)$ w.r.t. the line mirror $3 x + y = 6 a$ is the point $(a^{2} + 1, a)$ is