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Hello I think I know how to do my homework but I'm not sure.

Hello I think I know how to do my homework but I'm not sure.-example-1
User Oleg Sh
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4 votes

Answer:

The values of x and y for line PR and QS to be perpendicular are:


\begin{gathered} x=12 \\ y=23 \end{gathered}

Step-by-step explanation:

We want to find the value of x and y for which PR and QS are perpendicular.

For Line PR and line QS to be perpendicular;


\begin{gathered} \measuredangle PQS=90^0 \\ \measuredangle RQS=90^0 \end{gathered}

From the figure;


\measuredangle PQS=(4y-2)^0=90^0

solving for y;


\begin{gathered} 4y-2=90 \\ 4y=90+2 \\ 4y=92 \\ y=(92)/(4) \\ y=23 \end{gathered}

Also from the figure;


\measuredangle RQS=\measuredangle RQT+\measuredangle TQS=90^0

substituting the values;


\begin{gathered} \measuredangle RQS=\measuredangle RQT+\measuredangle TQS=90^0 \\ \measuredangle RQS=2x^0+(5x+6)^0=90^0 \\ \end{gathered}

Solving for x;


\begin{gathered} 2x+5x+6=90 \\ 7x+6=90 \\ 7x=90-6 \\ 7x=84 \\ \text{divide both sides by 7;} \\ (7x)/(7)=(84)/(7) \\ x=12 \end{gathered}

Therefore, the values of x and y for line PR and QS to be perpendicular are:


\begin{gathered} x=12 \\ y=23 \end{gathered}

User Delcon
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