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

y−coordinate is 2.

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=((x_1+x_2)/(2),(y_1+y_2)/(2))=((-1+1)/(2),(4-2)/(2))=(0,1)`

Slope of PQ is

`m_(PQ)=(y_2-y_1)/(x_2-x_1)=(-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)=(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)=(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=(|ax_1+by_1+c|)/(√(a^2+b^2))`

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

`√(10)=(|3a+b-1=0|)/(√(3^2+1^2))`

`√(10)=(|3a+b-1=0|)/(√(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.