Question

Find the point of the x-axis which is equidistant from the points (5, 4) and (-2, 3). Also find the area of triangle formed by these points

Answer 1

let the point P(x,0) which is equidistant from A(5,4) and B(-2,3)

now after applying distance formula to find AP, we have

AP = x^2-10x+41 under the root.........(1)

similarily, BP = x^2+4x+13 under the root................(2)

on comparing (1) and (2) we get:-

x^2-10x+41 = x^2+4x+13 (after squaring both sides)

10x+4x = 41-13

14x = 28

x = 2

so the point is P(2,0)

now after applying formula of area of triangle,

1/2|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|

=12.5sq.units

Answer 2

Answer:

let the point P(x,0) which is equidistant from A(5,4) and B(-2,3)

now after applying distance formula to find AP, we have

AP = x^2-10x+41 under the root.........(1)

similarily, BP = x^2+4x+13 under the root................(2)

on comparing (1) and (2) we get:-

x^2-10x+41 = x^2+4x+13 (after squaring both sides)

10x+4x = 41-13

14x = 28

x = 2

so the point is P(2,0)

now after applying formula of area of triangle,

1/2|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|

=12.5sq.units

Step-by-step explanation:

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