Question

Consecutive sides of a square are perpendicular; therefore, QR 1 SQ. Find thevalue of x.

Answer 1

SOLUTION

Since QR is perpendicular to SQ, then,

Let the grdient of the line QR be m1 and the gradient of the line SQ be m2.

Since QR is pendicular to SQ, m1 x m2 = -1

$\begin{gathered} \text{From m = }\frac{y2\text{ -y1}}{x2\text{ -x1}}\text{ } \\ m1\text{ = }\frac{1\text{ - 3}}{7\text{ - 3 }}\text{ = }\frac{-2}{4} \\ =\text{ }\frac{-1}{2} \end{gathered}$$\begin{gathered} m2\text{ = }\frac{3\text{ - }1}{3-\text{ x}}\text{ } \\ m2\text{ = }\frac{2}{3\text{ -x }} \end{gathered}$$\begin{gathered} \frac{-1}{2}\text{ }\times\frac{2}{3\text{ -x }}\text{ = -1 } \\ \frac{-2}{2(3\text{ -x)}}\text{ = -1} \\ \frac{-2}{6\text{ - 2x}}\text{ = -1 cross multiplying, we will have } \\ -1(6\text{ - 2x) = -2 } \\ -6\text{ + 2x = -2 } \\ 2x\text{ = -2 + 6} \\ 2x\text{ = 4 } \\ x\text{ = 2} \end{gathered}$

Therefore, x = 2