find a point on x axis which is at a distance of 2^5units from the point A(7,-4)
Answers
Answer:
Step-by-step explanation:
let coordinates of the point=(x,0)(as the poin lies on x axis)
x1=7. y1=-4
x2=x. y2=0
distance between 2 pts=√(x2-x1)^2+(y2-y1)^2
A/Q.....
2√5=√(x-7)^2+(0-(-4))^2
on squaring both sides....
20=x^2+49-14x+16
20=x^2+65-14x
0=x^2-14x+45
0=x^2-9x-5x+45
0=x(x-9)-5(x-9)
0=(x-9)(x-5)
x-9=0. x-5=0
x=9. x=5
therefore x=9,5
and coordinates of points.....(9,0)or(5,0)
let coordinates of the point=(x,0)(as the poin lies on x axis)
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2A/Q.....
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2A/Q.....2√5=√(x-7)^2+(0-(-4))^2
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2A/Q.....2√5=√(x-7)^2+(0-(-4))^2on squaring both sides....
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2A/Q.....2√5=√(x-7)^2+(0-(-4))^2on squaring both sides....20=x^2+49-14x+16
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2A/Q.....2√5=√(x-7)^2+(0-(-4))^2on squaring both sides....20=x^2+49-14x+1620=x^2+65-14x
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2A/Q.....2√5=√(x-7)^2+(0-(-4))^2on squaring both sides....20=x^2+49-14x+1620=x^2+65-14x0=x^2-14x+45
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2A/Q.....2√5=√(x-7)^2+(0-(-4))^2on squaring both sides....20=x^2+49-14x+1620=x^2+65-14x0=x^2-14x+450=x^2-9x-5x+45
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2A/Q.....2√5=√(x-7)^2+(0-(-4))^2on squaring both sides....20=x^2+49-14x+1620=x^2+65-14x0=x^2-14x+450=x^2-9x-5x+450=x(x-9)-5(x-9)
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2A/Q.....2√5=√(x-7)^2+(0-(-4))^2on squaring both sides....20=x^2+49-14x+1620=x^2+65-14x0=x^2-14x+450=x^2-9x-5x+450=x(x-9)-5(x-9)0=(x-9)(x-5)
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2A/Q.....2√5=√(x-7)^2+(0-(-4))^2on squaring both sides....20=x^2+49-14x+1620=x^2+65-14x0=x^2-14x+450=x^2-9x-5x+450=x(x-9)-5(x-9)0=(x-9)(x-5)x-9=0. x-5=0
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2A/Q.....2√5=√(x-7)^2+(0-(-4))^2on squaring both sides....20=x^2+49-14x+1620=x^2+65-14x0=x^2-14x+450=x^2-9x-5x+450=x(x-9)-5(x-9)0=(x-9)(x-5)x-9=0. x-5=0x=9. x=5
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2A/Q.....2√5=√(x-7)^2+(0-(-4))^2on squaring both sides....20=x^2+49-14x+1620=x^2+65-14x0=x^2-14x+450=x^2-9x-5x+450=x(x-9)-5(x-9)0=(x-9)(x-5)x-9=0. x-5=0x=9. x=5therefore x=9,5
let coordinates of the point=(x,0)(as the poin lies on x axis)x1=7. y1=-4x2=x. y2=0distance between 2 pts=√(x2-x1)^2+(y2-y1)^2A/Q.....2√5=√(x-7)^2+(0-(-4))^2on squaring both sides....20=x^2+49-14x+1620=x^2+65-14x0=x^2-14x+450=x^2-9x-5x+450=x(x-9)-5(x-9)0=(x-9)(x-5)x-9=0. x-5=0x=9. x=5therefore x=9,5and coordinates of points.....(9,0)or(5,0)