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Q1. Show that the point P(-4,2) lies on the line segment joining the points A(-4,6) and B (-4,-6).

Solution

We know, if points are collinear the area of the triangle formed by three points as vertices is zero. Area = [-4 (6 + 6) - 4 (-6 - 2) - 4(2 - 6)] = 0 So, the points are collinear.

Q2. Find a relation between x and y such that the point P(x,y) is equidistant from the points A(7,1) and B(3,5).

Solution

Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5). Given, AP = BP AP² = BP² By distance formula, (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)² x² + 49 - 14x + y² + 1 - 2y = x² + 9 - 6x + y²+ 25 - 10y -7x + 25 - y = -3x + 17 - 5y -4x + 8 + 4y = 0 x - y = 2

Q3. In which quadrant do the following points lie? A (-3, 5)          B (-2, -7)

Solution

In the point A (-3, 5) the abscissa is negative and ordinate is positive. So, it lies in the second quadrant. The point B (-2, -7) lies in the third quadrant because both the abscissa and the ordinate are negative.

Q4. If (-2, 2), (x, 8) and (6, y) are three con-cyclic points whose centre is (2, 5), then, find the values of x and y.

Solution

Let the given points be A(-2,2), B(x,8), C(6,y) and O(2,5). Then, OA=OB=OC (Radii of same circle) OA²=OB²=OC² (2+2)²+ (5-2)² = (2-x) ²+ (5-8)²­ = (2-6)²+ (5-y) ² 16+9=x²-4x+4+9 = 16+y²-10y+25   x²-4x-12=0 and y²-10y+16=0 (x-6)(x+2)=0 and (y-8) (y-2) =0 x=6,-2 and y=2, 8

Q5. Find the value of 's' if the point P(0,2) is equidistant from Q(3,s) and R(s,5).

Solution

PQ = PR PQ² = PR² (0 - 3)² + (2 - s)² =(0 - s)² + (2 - 5)² 9 + 4 - 4s + s² = s² + 9 4 = 4s s = 1