Question

Find the values of y for which the distance between the points P(2, -3) and Q(10,y) is units.

Answer 1

Answer:

If you have to points, (X1, Y1) and (X2,Y2), the way to find the distance between them is to make a right triangle. If you know that on any right triangle a^2 + b^2=c^2, we just need to find the two legs to find the hypotenuse. These two distances are x1-x2 and y1-y2. If you plug these values into the equation, and substitute c for d (distance), you get

(x1-x2)^2 + (y1-y2)^2=d^2

Answer 2

$\huge \bf{Solution.}$

Points P and Q are

P(2, –3) and Q(10,y)

PQ=10(given)

PQ²=10²=100

(10–2)²+{y–(–3)}²=100

(8)²+(y+3)²=100

64+y²+6y+9=100

y²+6y–27=0

Splitting the middle term 6y,

y²+9y–3y–27=0

y(y+9)–3(y+9)=0

(y+9)(y–3)=0

y+9=0 or y–3=0

y= –9 or y=3

y= –9, 3

Hence, the required value of y is–9 or 3.