Q3 of 53 Page 107

Find the point on the X-axis which is equidistant from A(-3, 4) and B(1, -4).

A point in the x = axis is of the form (a,0)

Distance d between two points(a,b) and (c,d)is given by



Distance between (-3,4) and (a,0) =


D =


D


Distance between (1,-4) and (a,0)


D =


D =


As the two points are equidistant from the point (a.0)


=


Squaring both sides, we get


(1-a)2 + 16 = (3 + a)2 + 16


1 + a2 -2a = 9 + a2 + 6a


8a = -8


a = -1


Hence the point is (-1,0)


More from this chapter

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1

Find the distance between each of the following pairs of points.

(1) A(2, 3), B(4, 1)


(2) P(-5, 7), Q(-1, 3)


(3) R(0, -3), S(0, 5/2)


(4) L(5, -8), M(-7, -3)


(5) T(-3, 6), R(9, -10)


(6) , X(11, 4)

2

Determine whether the points are collinear.

(1) A(1, -3), B(2, -5), C(-4, 7)


(2) L(-2, 3), M(1, -3), N(5, 4)


(3) R(0, 3), D(2, 1), S(3, -1)


(4) P(-2, 3), Q(1, 2), R(4, 1)

 4

Verify that points P(-2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle.

 5

Show that points P(2, -2), Q(7, 3), R(11, -1) and S (6, -6) are vertices of a parallelogram.