In each of the following find the value of ‘k’, for which th

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 Multiple Choice QuestionsShort Answer Type

131.

Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).

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132.

If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that  

AP space equals space 3 over 7 and P lies on the line segment Ab.

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 Multiple Choice QuestionsLong Answer Type

133.

Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.

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134.

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.

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 Multiple Choice QuestionsShort Answer Type

135.

Find the area of the triangle whose vertices are :
(2, 3), (–1, 0), (2, – 4)

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136.

Find the area of the triangle whose vertices are:
(–5, –1), (3, –5), (5, 2)

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137.

In each of the following find the value of ‘k’, for which the points are collinear
(7, –2), (5, 1), (3, k)


Let the given points be A(7, -2), B(5, 1) and C(3, K)
Here, we have,
x1 = 7, y1 = -2
x2 = 5, y2 = 1
and    x3 = 3, y3 = K
Now, Area of ∆ABC

equals 1 half left square bracket straight x subscript 1 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 plus straight x subscript 3 left parenthesis straight y subscript 1 minus straight y subscript 2 right parenthesis right square bracket
equals 1 half left square bracket 7 left parenthesis 1 minus straight k right parenthesis plus 5 left curly bracket straight k minus left parenthesis negative 2 right parenthesis right curly bracket plus 3 left parenthesis negative 2 minus 1 right parenthesis right square bracket
equals 1 half left square bracket 7 minus 7 straight k plus 5 straight k plus 10 minus 9 right square bracket
equals 1 half left square bracket 8 minus 2 straight k right square bracket equals 4 minus straight k

If the points are collinear, then area of the triangle = 0
⇒    4 - k = 0
⇒    k = 4.

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138.

In each of the following find the value of ‘k’, for which the points are collinear
(8, 1), (k, – 4), (2, –5) 

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139.

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

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 Multiple Choice QuestionsLong Answer Type

140.

Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

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