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Two points A and B with $(1,1)$ and $(-2,3)$ respectively are given.find the locus of point P.So that area of $\Delta$PAB is 9 square units.
answer is :- $2x+3y+13=0$ or $2x+3y-23=0$.
how i tried:- i assigned unknown point to be P$(x,y)$ and completed the triangle and from point P i drew a perpendicular until point O(a random point on line-segment AB).

enter image description here i found point O in terms of $x_1$ and $y_1$ using equation of line.Now perpendicular line PO is height and the base is AB. Now i used formula $\dfrac{1}{2}(base *height)$.On equating the following this equation with 9, i ended up with the wrong answer.please can someone help me find the solution?? please.

PVAL-inactive
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Nikhil01
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1 Answers1

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Assuming $P(h,k),$ I found $$\frac12\cdot\sqrt{13}\cdot\frac{|2h+3k-5|}{\sqrt{13}}=9\iff|2h+3k-5|=18$$

which complies with the given answer

lab bhattacharjee
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  • how did you get the above answer sir?? – Nikhil01 Jul 27 '14 at 13:07
  • @Nikhil01, $|AB|=\cdots=\sqrt{13}$ unit, right? What is the perpendicular distance of $AB:2x+3y-5=0$ from $P(h,k)?$ – lab bhattacharjee Jul 27 '14 at 13:09
  • i'm really sorry disturbing you sir, but i do not understand how did you find the eqution $2x+3y-5=0$and the perpendicular distance.your help would be grateful. – Nikhil01 Jul 27 '14 at 13:23
  • @Nikhil01, $$\frac{y-1}{x-1}=\frac{1-3}{1+2}$$ and http://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line#Cartesian_coordinates – lab bhattacharjee Jul 27 '14 at 13:26
  • I think it would be useful if @Nikhil01 would show the calculation that led to the wrong result, so that others can try to spot the specific mistake. The description of the method, as given in the question, looks OK to me. But it says "using equation of line", and Nikhil01's latest comment indicates that he doesn't know how to find the equation of the line AB. So it's not clear to me what his calculation actually did. – Andreas Blass Jul 27 '14 at 15:25
  • how i have done:- $y_1=mx_1+c$ on calculating, $m=\dfrac{2}{-3}$ and value of y-intercept $c=\dfrac{5}{3}$(found by substituting point A$(1,1)$and m in equation $y_1=mx_1+c$).On doing so i get following equation:-$y_1=\dfrac{2}{-3}x_1+\dfrac{5}{3}$.Here is where i went wrong :- i reassigned point O$(x_1,y_1)$ as $(x_1,\dfrac{2}{-3}x_1+\dfrac{5}{3})$ and tried to find the distance of line segment PO using distance formula.I should have used this formula to find the distance of PO :-$\dfrac{|ax+by+c|}{\sqrt{a^2+b^2}}$.n thankx for your help guys – Nikhil01 Jul 28 '14 at 07:44
  • thank you lab bhattacharjee and Andreas Blass.I got the answer i have been looking for. – Nikhil01 Jul 28 '14 at 07:52