Ex 10.1, 4
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Let the given points be
A(7, 6) & B (3, 4)
Let C be a point on the x-axis
Coordinates of C = C(x, 0)
Given that point C is
equidistant from the points A & B
(As it is on the x-axis, y = 0)
Hence,
Distance AC = Distance BC
We know that
distance D between two points (x1 , y1)&(x2 , y2)is
D = √((𝑥2−𝑥1)2+(𝑦2 − 𝑦1)2)
Distance between A(7, 6) & C(x, 0)
AC = √((𝑥−7)^2+(0−6)2)
= √((𝑥−7)^2+36)
Distance between B(3, 4) & C(x, 0)
BC = √((𝑥−3)^2+(0−4)2)
= √((𝑥−3)^2+16)
Since,
AC = BC
√((𝑥−7)^2+36) = √((𝑥−3)^2+16)
Squaring both sides
(√((𝑥−7)^2+36))^2 = (√((𝑥−3)^2+16))^2
(x – 7)2 + 36 = (x – 3)2 + 16
(x – 7)2 – (x – 3)2 = 16 – 36
x2 + 49 – 14x – (x2 + 9 – 6x) = –20
x2 + 49 – 14x – x2 – 9 + 6x = –20
0 – 8x + 40 = −20
–8x = – 20 – 40
–8x = −60
x = (−60)/(−8)
x = 15/2
Thus, Required point = C(x, 0) = (𝟏𝟓/𝟐 ", 0" )

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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