find the points on x axis from which (7,6)and (-3,4)are equidistant
Answers
Answer:
A(7,6). B(-3,0)
C(x,0)
AC = BC
=>(7-X)²+ (6-0)² = (-3-X)² + (0)²
=>49+X²- 14X + 36 = 9+X² +6X
=> X² FROM BOTH SIDES WILL BE ELIMINATED
=> 85-14x = 9 + 6x
=> 76 = 20x
=> x = 4 (approx)
ur ans is (4,0)
Here is your answer
Here is your answerHorizontal lines have a slope of 0. Thus, in the slope-intercept equation y = mx + b, m = 0. The equation becomes y = b, where b is the y-coordinate of the y-intercept.Here is your answer
Here is your answerHorizontal lines have a slope of 0. Thus, in the slope-intercept equation y = mx + b, m = 0. The equation becomes y = b, where b is the y-coordinate of the y-intercept.Here is your answerHorizontal lines have a slope of 0. Thus, in the slope-intercept equation y = mx + b, m = 0. The equation becomes y = b, where b is the y-coordinate of the y-intercept.
Answer:
Answer: (3,0)
Step-by-step explanation:
We know that y-co-ordinate of a point on X-axis is always 0.
So, let a point on X-axis be P(x,0) and let two given points be A(7,6)andB(−3,4).
According to the condition,
PA=PB
⇒ √(x-7)²+(0-6)² = √(x+3)²+(0-4)²
Squaring both sides, we have
x²−14x+49+36 = x²+6x+9+16
∴ Required point is (3, 0)