distance between the points (7,4) and (3,y) is 4 units find the value of y.. pls its urgent i will mark it as brainliest... i want it with steps plsssssssssssssssssssssssssssssssssssss.
Answers & Comments
mahijiwani
Let p(7, 4) =(x1, y1) Q(3, y) =(x2, y2) PQ=4 units By distance formula, PQ=√(x1-x2) ^2 + (y1-y2) ^2 4=√(7-3) ^2 +(4-y) ^2 squaring on both sides, 4^2=(4) ^2 +(4-y) ^2 16=16+(4) ^2-2*4y+y^2 16=16+16-8y+y^2 16=32-8y+y^2 y^2-8y+32-16=0 y^2-8y+16=0 y^2-4y-4y+16=0 y(y-4) -4(y-4) =0 (y-4) (y-4) =0 y-4=0 y=4
Answers & Comments
Q(3, y) =(x2, y2)
PQ=4 units
By distance formula,
PQ=√(x1-x2) ^2 + (y1-y2) ^2
4=√(7-3) ^2 +(4-y) ^2
squaring on both sides,
4^2=(4) ^2 +(4-y) ^2
16=16+(4) ^2-2*4y+y^2
16=16+16-8y+y^2
16=32-8y+y^2
y^2-8y+32-16=0
y^2-8y+16=0
y^2-4y-4y+16=0
y(y-4) -4(y-4) =0
(y-4) (y-4) =0
y-4=0
y=4