Question

The side PQ of a square PQRS is parallel to Y-axis. Find the slopes of PS, PR and QS. <br /><br /><br />plz help me... ​

Answer 1

the slopes of PS is 0 , PR is -1 and QS is 1

• The slope of a line is tanθ where θ is the angle that the line makes with the x-axis.
• The line PS makes 0° with the x-axis so the slope will be tan0°=0
• The line PR makes 135° with the x-axis so the slope will be tan135°=-1
• The line QS make 45° with the x-axis so the slope will be tan45°=1

Answer 2

Given :

PQRS is a square with QS and PR as it's diagonals.

Side PQ is parallel to y-axis.

To Find :

Slopes of the segments PS, PR and QS

Solution:

1). Slope of side PS which parallel to x-axis = 0

  Since slope of x axis = 0 and any line parallel to the x-axis will have         the same slope.

2). Since diagonal PR is inclined at 135° to the x-axis,

   Therefore, slope of the segment PR ($m_1$) = tan(135)°

   $m_1$ = -1

3). Let the slope of another diagonal of PQRS is = $m_2$

   By the property of perpendicular liens,

   $m_1\times m_2=(-1)$

   $(-1)\times (m_2)=(-1)$

   $m_2=1$

  Therefore, slope of diagonal QS = 1