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If P(6,-2). O(-2,8), R(-4, 3), and S(-9, y). find the value of y so that PO perpendicular to RS.please?

User Obenland
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1 Answer

3 votes

Answer:

y = - 1

Step-by-step explanation:

Two lines are perpendicular if the product of their slopes is equal to -1.

Additionally, we can calculate the slope of a line with two points (x1, y1) and (x2, y2) as:


m=(y_2-y_1)/(x_2-x_1)

If we replace (x1, y1) by P(6, -2) and (x2, y2) by O(-2, 8), we get that the slope of PO is equal to:


m=(8-(-2))/(-2-6)=(8+2)/(-8)=(10)/(-8)=-1.25

In the same way, if we replace (x1, y1) by (-4, 3) and (x2, y2) by (-9, y), we get that the slope of RS is equal to:


m_{}=(y-3)/(-9-(-4))=(y-3)/(-9+4)=(y-3)/(-5)

Then, the product of these two slopes should be equal to -1, so we can write the following equation:


-1.25\cdot((y-3)/(-5))=-1

So, solving for y, we get:


\begin{gathered} (-5)(-1.25)\cdot((y-3)/(-5))=(-5)(-1) \\ -1.25(y-3)=5 \\ y-3=(5)/(-1.25) \\ y-3=-4 \\ y=-4+3 \\ y=-1 \end{gathered}

Therefore, the value of y is equal to -1

User JimDusseau
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