Answer :
Answer:
y−coordinate is 2.
Step-by-step explanation:
It is given that P(−1,4),Q(1,−2), with K a point on the perpendicular bisector of PQ.
K is in the first quadrant at a distance of √10 units from PQ.
Let the coordinates of K are (a,b) where a and b are non negative.
A be the midpoint of PQ is
[tex]A=(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})=(\dfrac{-1+1}{2},\dfrac{4-2}{2})=(0,1)[/tex]
Slope of PQ is
[tex]m_{PQ}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-2-4}{1-(-1)}=-3[/tex]
PQ and AK are perpendicular. It means Product of their slopes is -1.
[tex]m_{PQ}\cdot m_{KA}=-1[/tex]
[tex]-3\cdot m_{KA}=-1[/tex]
[tex]\cdot m_{KA}=\frac{1}{3}[/tex]
Point slope form of a line is
[tex](y-y_1)=m(x-x_1)[/tex]
where, m is slope.
Equation PQ is
[tex](y-4)=-3(x-(-1))[/tex]
[tex]y-4=-3x-3[/tex]
[tex]3x+y-4+3=0[/tex]
[tex]3x+y-1=0[/tex]
Similarly. equation of AK is
[tex](y-0)=\frac{1}{3}(x-1)[/tex]
[tex]x-3y=-3[/tex]
It passes through (a,b) so
[tex]a-3b=-3[/tex] ... (1)
The distance of a point [tex](x_1,y_1)[/tex] from the line [tex]ax+by+x=0[/tex] is
[tex]d=\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}[/tex]
It is given that the distance between K and PQ is √10.
[tex]\sqrt{10}=\dfrac{|3a+b-1=0|}{\sqrt{3^2+1^2}}[/tex]
[tex]\sqrt{10}=\dfrac{|3a+b-1=0|}{\sqrt{10}}[/tex]
[tex]10=|3a+b-1=0|[/tex]
[tex]\pm 10=3a+b-1=0[/tex]
[tex]- 10=3a+b-1=0\rightarrow 3a+b=-9[/tex] .... (2)
[tex]10=3a+b-1=0\Rightarrow 3a+b=11[/tex] ... (3)
On solving (1) and (2) we get
[tex]a=-3,b=0[/tex]
On solving (1) and (3) we get
[tex]a=3,b=2[/tex]
We know that K is in the first quadrant. It means a≥0 and b≥0.
Therefore, y−coordinate is 2.
