A line having slope '1' is drawn from a point $A(-3,0)$ cuts a curve $y=x^2+x+1$ at $P \,\&\, Q$. Find $\ell({AP}) \ell({AQ})$.

JEE MATHEMATICS MCQ: STRAIGHT LINES

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Application +4 | -1

A line having slope '1' is drawn from a point

A(-3,0)

cuts a curve

y=x^2+x+1

P \,\&\, Q

. Find

\ell({AP}) \ell({AQ})

View answer and explanation

Correct option: A

Option A (14) is correct because it is the product of the distances from the point A to the intersection points P and Q, found by solving the quadratic equation and applying Vieta's formulas.

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x-3 y=p, a x+2 y=q

and

a x+y=r

form a right-angled triangle then :
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\left\{ {a{t_1}{t_2},\;a\left( {{t_1} + {t_2})\} ,\{ a{t_2}{t_3},a({t_2} + {t_3}} \right)} \right\},\;\left\{ {a{t_3}{t_1},\;a\left( {{t_3} + {t_1}} \right)} \right\}

is
The distance between the parallel lines

y = 2\;x + 4

and

6x = 3y + 5

is The distance between the pair of lines given by

{x^2} + {y^2} + 2xy - 8ax - 8ay - 9{a^2} = 0

is Find the image of the point

(3,8)

with respect of the line

x+3 y=7

assuming the line to be plane mirror: The centroid of the triangle whose three sides are given by the combined equation

\left( {{x^2} + 7xy + 2{y^2}} \right)\left( {y - 1} \right) = 0