Coordinate Geometry

  • Question 1
    CBSEENMA10009647

    If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k?

    Solution

    Let the points X (4,k) and Y(1,0)
    It is given that the distance XY is 5 units.
    By using the distance formula,
    XY space equals space square root of left parenthesis 4 minus 1 right parenthesis squared space plus left parenthesis straight k minus 0 right parenthesis squared end root
rightwards double arrow space 5 space equals space square root of 9 plus left parenthesis straight k right parenthesis squared end root
rightwards double arrow space 25 space equals space 9 space plus straight k squared
rightwards double arrow 16 space equals space straight k squared
rightwards double arrow plus-or-minus space 4 space equals space straight k

    Question 2
    CBSEENMA10009662

    The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, –2). If the third vertex is (7/2, y), find the value of y.

    Solution

     Let A(x1, y1)=A(2, 1), B(x2, y2)=B(3,−2) and C(x3, y3)=C(7/2,y).
    Area space left parenthesis increment ABC right parenthesis space equals space 1 half vertical line straight x subscript 1 space left parenthesis straight y subscript 2 minus straight y subscript 3 right parenthesis plus straight x subscript 2 left parenthesis straight y subscript 3 minus straight y subscript 1 right parenthesis space plus straight x subscript 3 left parenthesis straight y subscript 1 minus straight y subscript 2 right parenthesis vertical line
rightwards double arrow space 5 space equals space 1 half vertical line 2 left parenthesis negative 2 minus straight y right parenthesis plus 3 left parenthesis straight y minus 1 right parenthesis space plus 7 over 2 left parenthesis 1 plus 2 right parenthesis vertical line
rightwards double arrow 10 space equals space open vertical bar negative 4 minus 2 straight y plus 3 straight y minus 3 space plus 21 over 2 close vertical bar
rightwards double arrow space 10 space equals space open vertical bar straight y plus 7 over 2 close vertical bar
rightwards double arrow space 10 space equals space straight y space plus 7 over 2
or minus 10 space equals space straight y space plus 7 over 2
rightwards double arrow space straight y space equals space 13 over 2 space or space straight y space equals space fraction numerator negative 27 over denominator 2 end fraction

    Question 3
    CBSEENMA10009672

    If a≠b≠0, prove that the points (a, a2), (b, b2) (0, 0) will not be collinear.

    Solution

    Let A(a, a2), B(b, b2) and C(0, 0) be the coordinates of the given points.
    We know that the area of a triangle having vertices (x1, y1), (x2, y2) and (x3, y3) is ∣∣12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣∣ square units.
    So,
    Area of ∆ABC\
    open vertical bar 1 half open square brackets straight a left parenthesis straight b squared minus 0 right parenthesis plus straight b left parenthesis 0 minus straight a squared right parenthesis plus 0 left parenthesis straight a squared minus straight b squared right parenthesis close square brackets close vertical bar
space equals space open vertical bar 1 half left parenthesis ab squared minus straight a squared straight b right parenthesis close vertical bar
space equals 1 half open vertical bar ab left parenthesis straight b minus straight a right parenthesis close vertical bar
not equal to 0 space left parenthesis therefore straight a not equal to straight b not equal to 0 right parenthesis
    Since the area of the triangle formed by the points (a, a2), (b, b2) and (0, 0) is not zero, so the given points are not collinear.

    Question 4
    CBSEENMA10009676

    In what ratio does the x-y-2=0 divided the line segment joining the points A(3,-1) and B(8,9)?

    Solution

    let t line x-y-2=0 divid t line segment in t ratio k : 1 at point C
    so coordinates of C are

    open parentheses fraction numerator 8 straight k plus 3 over denominator straight k plus 1 end fraction close parentheses comma open parentheses fraction numerator 9 straight k minus 1 over denominator straight k plus 1 end fraction close parentheses

    C lies on x - y - 2 = o

    so, subs coordinates of x,y in tis eq.
    open parentheses fraction numerator 8 straight k plus 3 over denominator straight k plus 1 end fraction close parentheses minus open parentheses fraction numerator 9 straight k minus 1 over denominator straight k plus 1 end fraction close parentheses space minus 2 space equals 0
fraction numerator 8 k plus 3 minus 9 k plus 1 minus 2 k minus 2 over denominator k plus 1 end fraction space equals 0
fraction numerator 8 k plus 3 minus 9 k plus 1 minus 2 k minus 2 over denominator k plus 1 end fraction space equals space 0
fraction numerator negative 3 k plus 2 over denominator k plus 1 end fraction space equals space 0
minus 3 k space plus 2 space equals space 0
rightwards double arrow space k space equals space 2 divided by 3
i. e space 2 colon 3

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