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Math Help - Slopes and graph questions

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    Thumbs down Slopes and graph questions

    Find the direction in which a straight line must be drawn through the point (-1,2) so that it's point of intersection with the line x + y = 4 may be at a distance of 3 units from this point. Oh yeah I used graph , I'm getting a line parallel line to x- axis... How o solve this analyticaly...?

    Can't solve this question du to too many unknown variables.... besides I dont think they are asking for perpendicular distance, be cause when I checked with the formula d = abs ( Ax1 + By1 + C)/SQRT(A^2 B^2)'whrer x1 = -1 , y1 = 2 ...

    Q2 The hypotenuse of a right trianlge has its ends at the points (1,3) and (-4,1). Find te equaion of the legs ( perpendicular sides ) of this triangle.

    I don't think there is a solution for both the legs because I feel that the hypotenuse can be considered asa diamter of a cirlce , and it subtends 90 degree at infinite points on it's circumference, but my textbook says the anser is x=1 , y=1 ... shouldnt we get a quadratic equation ??


    Ice Sync
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    Quote Originally Posted by ice_syncer View Post
    Find the direction in which a straight line must be drawn through the point (-1,2) so that it's point of intersection with the line x + y = 4 may be at a distance of 3 units from this point. Oh yeah I used graph , I'm getting a line parallel line to x- axis... How o solve this analyticaly...?

    Can't solve this question du to too many unknown variables.... besides I dont think they are asking for perpendicular distance, be cause when I checked with the formula d = abs ( Ax1 + By1 + C)/SQRT(A^2 B^2)'whrer x1 = -1 , y1 = 2 ...

    Q2 The hypotenuse of a right trianlge has its ends at the points (1,3) and (-4,1). Find te equaion of the legs ( perpendicular sides ) of this triangle.

    I don't think there is a solution for both the legs because I feel that the hypotenuse can be considered asa diamter of a cirlce , and it subtends 90 degree at infinite points on it's circumference, but my textbook says the anser is x=1 , y=1 ... shouldnt we get a quadratic equation ??


    Ice Sync

    So the equatio of a circle with radius 3 centered at the point (-1,2) is

    (x+1)^2+(y-2)^2=9 and we want to find its point of intersection with the line x+y=4 \iff x=4-y

    So if we sub into the equation for the circle and factor we get

    (5-y)^2+(y-2)^2=9 \iff 2y^2-14y+20=0 \iff 2(y-2)(y-5)=0

    So we get y=2 \mbox{ or } y=5

    Subbing each of these into the equation of the line we complete the ordered pairs to get the points

    (2,2) and (-1,5)
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    Quote Originally Posted by TheEmptySet View Post
    So the equatio of a circle with radius 3 centered at the point (-1,2) is

    (x+1)^2+(y-2)^2=9 and we want to find its point of intersection with the line x+y=4 \iff x=4-y

    So if we sub into the equation for the circle and factor we get

    (5-y)^2+(y-2)^2=9 \iff 2y^2-14y+20=0 \iff 2(y-2)(y-5)=0

    So we get y=2 \mbox{ or } y=5

    Subbing each of these into the equation of the line we complete the ordered pairs to get the points

    (2,2) and (-1,5)


    Null Set,

    well, nice idea, we weren't taught cirlce equation and all, but the qustion was.. what direction, not point of intersection.. so is the direction tan(135) ??
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    Quote Originally Posted by ice_syncer View Post
    Null Set,

    well, nice idea, we weren't taught cirlce equation and all, but the qustion was.. what direction, not point of intersection.. so is the direction tan(135) ??

    If you know points of intersection how can you answer your question?

    Hint: if you start at (-1,2) how do you get to the point (2,2).

    If you cant see it draw a graph, and your answer is incorrect.

    P.S if you use the distance forumla

    d=\sqrt{(x+1)^2+(y-2)^2} and set it equal to 3 and square it you get

    3=\sqrt{(x+1)^2+(y-2)^2} \iff 9=(x+1)^2+(y-2)^2

    And the solution follows just as above
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