Question

Given P(−1,4),Q(1,−2), with K a point on the perpendicular bisector of PQ. If K is in the first quadrant at a distance of √10 units from PQ, what is its y−coordinate?

Answer 1

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

$A=(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})=(\dfrac{-1+1}{2},\dfrac{4-2}{2})=(0,1)$

Slope of PQ is

$m_{PQ}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-2-4}{1-(-1)}=-3$

PQ and AK are perpendicular. It means Product of their slopes is -1.

$m_{PQ}\cdot m_{KA}=-1$

$-3\cdot m_{KA}=-1$

$\cdot m_{KA}=\frac{1}{3}$

Point slope form of a line is

$(y-y_1)=m(x-x_1)$

where, m is slope.

Equation PQ is

$(y-4)=-3(x-(-1))$

$y-4=-3x-3$

$3x+y-4+3=0$

$3x+y-1=0$

Similarly. equation of AK is

$(y-0)=\frac{1}{3}(x-1)$

$x-3y=-3$

It passes through (a,b) so

$a-3b=-3$          ... (1)

The distance of a point $(x_1,y_1)$ from the line $ax+by+x=0$ is

$d=\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$

It is given that the distance between K and PQ is √10.

$\sqrt{10}=\dfrac{|3a+b-1=0|}{\sqrt{3^2+1^2}}$

$\sqrt{10}=\dfrac{|3a+b-1=0|}{\sqrt{10}}$

$10=|3a+b-1=0|$

$\pm 10=3a+b-1=0$

$- 10=3a+b-1=0\rightarrow 3a+b=-9$       .... (2)

$10=3a+b-1=0\Rightarrow 3a+b=11$          ... (3)

On solving (1) and (2) we get

$a=-3,b=0$

On solving (1) and (3) we get

$a=3,b=2$

We know that  K is in the first quadrant. It means a≥0 and b≥0.

Therefore, y−coordinate is 2.