KnowledgeBoat Logo
|

Mathematics

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

Coordinate Geometry

3 Likes

Answer

By formula,

Distance between two points (D) = (y2−y1)2+(x2−x1)2\sqrt{(y2 - y1)^2 + (x2 - x1)^2}

Substituting values we get :

⇒10=[y−(−3)]2+(10−2)2⇒10=[y+3]2+82⇒10=y2+9+6y+64⇒10=y2+6y+73⇒102=(y2+6y+73)2⇒100=y2+6y+73⇒y2+6y+73−100=0⇒y2+6y−27=0⇒y2+9y−3y−27=0⇒y(y+9)−3(y+9)=0⇒(y−3)(y+9)=0⇒y−3=0 or y+9=0⇒y=3 or y=−9.\Rightarrow 10 = \sqrt{[y - (-3)]^2 + (10 - 2)^2} \\[1em] \Rightarrow 10 = \sqrt{[y + 3]^2 + 8^2} \\[1em] \Rightarrow 10 = \sqrt{y^2 + 9 + 6y + 64} \\[1em] \Rightarrow 10 = \sqrt{y^2 + 6y + 73} \\[1em] \Rightarrow 10^2 = \Big(\sqrt{y^2 + 6y + 73}\Big)^2 \\[1em] \Rightarrow 100 = y^2 + 6y + 73 \\[1em] \Rightarrow y^2 + 6y + 73 - 100 = 0 \\[1em] \Rightarrow y^2 + 6y - 27 = 0 \\[1em] \Rightarrow y^2 + 9y - 3y - 27 = 0 \\[1em] \Rightarrow y(y + 9) - 3(y + 9) = 0 \\[1em] \Rightarrow (y - 3)(y + 9) = 0 \\[1em] \Rightarrow y - 3 = 0 \text{ or } y + 9 = 0 \\[1em] \Rightarrow y = 3 \text{ or } y = -9.

Hence, y = 3 or -9.

Answered By

2 Likes

Related Questions