Question

squidistant from (2,-) and (2,9).
Find the values of y for which the distance between the points P(2,-- 3) and Q(10,y) is
10 units.​

Answer 1

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

Answer :--

Formula used :- Distance between two points P(x1,y1) and Q(x2,y2) is given by: D(PQ) = √[(x2 - x1)² + (y2 - y1)²]

we have :-

→ P(2,-3) = x1 = 2 , y1 = (-3)

→ Q(10,y) = x2 = 10, y2 = y

→ D = 10 units .

Putting all values in above told formula now, we get,

→ 10 = √[(10-2)² + (y-(-3))²]

→ 10 = √[8² + (y+3)²]

Squaring both sides now,

→ 100 = 64 + y² + 9 + 6y

→ y² + 6y -27 = 0

Splitting the Middle term now,

→ y² + 9y - 3y - 27 = 0

→ y(y+9) -3(y+9) = 0

→ (y+9)(y-3) = 0

Putting both Equal to Zero now, we get,

→ y + 9 = 0

→ y = (-9)

or,

→ y-3 = 0

→ y = 3 .

Hence, the value of y is (-9) and (3) both Possible .

Answer 2

Correct Question:

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

Solution:

Given:

• Distance between points P and Q = 10 units.
• Point P = (2, -3)
• Point Q = (10, y)

To find:

• Value of y.

Now, by using Distance formula,

$\implies \sf Distance = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}};\;\;\;\;[Distance\;Formula]\\ \\ \\ \sf Where,\;x_{1}=2,\;x_{2}=10,\;y_{1}=-3\;and\; y_{2}=y.\\ \\ \\ \implies \sf 10=\sqrt{(10-2)^{2}+[y-(-3)]^{2}}\\ \\ \\ \implies \sf 10=\sqrt{(8)^{2}+(y+3)^{2}}\\ \\ \\ \implies \sf 10 = \sqrt{64+y^{2}+6y+9}\\ \\ \\ {\underline{\bf Now,\;squaring\;both\;the\;sides,}}\\ \\ \\ \implies \sf 100 = 64+y^{2}+6y+9 \\ \\ \\ \implies \sf 100 = 73+y^{2}+6y\\ \\ \\ \implies \sf 27=y^{2}+6y$

$\implies \sf y^{2} +6y-27=0\\ \\ \\ {\underline{\bf By\;using\;splitting\;middle\;term,}}\\ \\ \\ \implies \sf y^{2}+9y-3y-27=0\\ \\ \\ \implies y(y+9)-3(y+9)=0\\ \\ \\ \implies (y-3)(y+9)=0\\ \\ \\ \implies \sf y = 3\;and\; -9\\ \\ \\ {\underline{\bf Hence,\;value\;of\;y=3\;and\;-9.}}$