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)

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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) 


Let the given points be A(8, 1), B(K, -4) and C(2, -5).
Here, we have
x1 = 8, y1 = 1
x2 = K, y2 = -4
and    x3 = 2, y3 = -5
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 3 right square bracket
equals 1 half left square bracket 8 left curly bracket negative 4 minus left parenthesis negative 5 right parenthesis right curly bracket plus straight k left parenthesis negative 5 minus 1 right parenthesis plus 2 left curly bracket 1 minus left parenthesis negative 4 right parenthesis right curly bracket right square bracket
equals 1 half left square bracket 8 minus 6 straight k plus 10 right square bracket
1 half left square bracket 18 minus 6 straight k right square bracket equals 9 minus 3 straight k

If the points are collinear, then area of the triangle = 0

rightwards double arrow space 9 minus 3 straight k equals 0
rightwards double arrow space 3 straight k space equals space 9
rightwards double arrow space straight k space equals space 9 over 3 equals 3


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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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