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`(|ax_(1)^(2)+2hx_(1)y_(1)+by_(1)^(2)|)/(sqrt((a-b)^(2)+4h^(2)))``(|ax_(1)^(2)+2hx_(1)y_(1)+by_(1)^(2)|)/(sqrt((a-b)^(2)+h^(2)))``(|ax_(1)^(2)+2hx_(1)y_(1)+by_(1)^(2)|)/(sqrt((a+b)^(2)+4h^(2)))``(|ax_(1)^(2)-2hxy_(1)y_(1)+by_(1)^(2)|)/(sqrt((a-b)^(2)+4h^(2)))`

Answer :

ASolution :

Let `y-x_(1)x=0` and `y=m_(2)x=0` be the lines represented by the equation `ax^(2)+2hxy+by^(2)=0`. Then, <br> `m_(1)+m_(2)=-(2h)/(b) and m_(1)m_(2)=(a)/(b)" (i)"` <br> Let `p_(1) and p_(2)` be the lengths of perpendiculars from `(x_(1),y_(1))` on lines `y-m_(1)x=0` and `y-m_(2)x=0`. Then, <br> `p_(1)=(y_(1)-m_(1)x_(1))/(sqrt(1+m_(1)^(2)))andp_(2)=(y_(1)-m_(2)x_(1))/(sqrt(1+x_(2)^(2)))` <br> `therefore" "p_(1)p_(2)=((y_(1)-m_(1)x_(1))/(sqrt(1+x_(1)^(2))))((y_(1)-m_(2)x_(1))/(sqrt(1+m_(2)^(2))))=((y_(1)-m_(1)x_(1))(y_(1)-m_(2)x_(1)))/(sqrt((1+m_(1)^(2))(1+m_(2))))` <br> `rArr" "p_(1)p_(2)=(y_(1)^(2)-x_(1)y_(1)(m_(1)+m_(2))+m_(1)m_(2)x_(1)^(2))/(sqrt(1+(m_(1)^(2)+m_(2)^(2))+m_(1)^(2)m_(2)^(2)))` <br> `rArr" "p_(1)p_(2)=(y_(1)^(2)-x_(1)y_(1)(m_(1)+m_(2))+m_(1)m_(2)x_(1)^(2))/(sqrt(1+(m_(1)+m_(2))^(2)-2m_(1)m_(2)+(m_(1)m_(2))^(2)))` <br> `rArr" "p_(1)p_(2)=(y_(1)^(2)-x_(1)y_(1)(-(2h)/(b))+(a)/(b)x_(1)^(2))/(sqrt(1+(4h^(2))/(b^(2))-(2a)/(b)+(a^(2))/(b^(2))))" [Using (i)]"` <br> `rArr" "p_(1)p_(2)=(ax_(1)^(2)+2hx_(1)y_(1)+by_(1)^(2))/(sqrt((a-b)^(2)+4h^(2)))`Transcript

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00:00 - 00:59 | today special is the product of perpendicular let fall from X1 Y1 upon the lines represented by X square + 2 X square + Y square is equal to zero if we have the blind as X square + 2 X + b y square is equal to now dividing the whole equation by X where we get a square B square + 2 x y by x is equal to the slope of any line can be represented by a is equal to y by x to changing Vibe X2 and we get BM square + 2 X + Y is equal to zero now let the two lines represented by this equation let Y - 1 x is equal to zero |

01:00 - 01:59 | and Y - M 2x is equal to zero sum of the slow can be represented by this equation is quadratic in which is vichar clue to root 6 + sum of roots can be represented by -2 HV and product can be represented by M1 M2 that is equal to now we know that if we have a given point ab and equation of line given is PX + Q + security to the perpendicular distance of a we can be given by x is equal to a + b + b square + c square |

02:00 - 02:59 | at the given equation Y - 1 let the positive 2 - 1 X 1 by root of 1 + a square NP 27 - 2 X 1 by root over of oneplus m to the product of perpendicular product of perpendicular 20 P1 and P2 which is equal to 1 minus 11 x 17 - 2 X 1 divided by root 2 + M1 M2 and two square opening a |

03:00 - 03:59 | bracket we get into P2 equal to mode of square minus A + M2 / root over of 1 + 1 square + 2 square + 1 into m24 now we are having sum of roots as -2 h bi bi the changes to + X + x square divided by X square + a square can be replaced by 1 + 1 + M2 whole square minus to Eminem |

04:00 - 04:59 | love substituting MN + N2 and M1 M2 by their value we get a square + b + c y b by square by b square minus to know if we see we are having 1 -2 see why we are having a live show if we see we're having 1 -2 ABB + a square b square so this is the |

05:00 - 05:59 | formula of 1 minus b whole square square + 2 X 1 + x square by root 1 minus we are taking one by be performed so we get one by we would be minus plus which food we get a b y square + x square Baran to karo B minus a whole square + 2 square if we see in the |

06:00 - 06:59 | ocean we have option air |

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