The line through the points p(-3,-2) and Q(2,3) is perpendicular to the line through points R(10,-1) and S(15,-6) because the product of their slopes is -1
C: Yes, their slope have product -1
Two lines are perpendicular if the products of their slope equals -1
The slope(m) of a line passing through the points [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] is given as:
[tex]m = \frac{y_2-y_1}{x_2-x_1}[/tex]
For the line PQ:
P(-3, -2) and Q(2, 3)
[tex]m_1 = \frac{3-(-2)}{2-(-3)} \\\\m_1 = \frac{3+2}{2+3} \\\\m_1 = \frac{5}{5} \\\\m_1 = 1[/tex]
For the line RS:
R(10,-1) and S(15,-6)
[tex]m_2 = \frac{-6-(-1)}{15-10} \\\\m_2 = \frac{-6+1}{5} \\\\m_2 = \frac{-5}{5} \\\\m_2 = -1[/tex]
[tex]m_1m_2 = -1(1) \\\\m_1m_2 = -1[/tex]
Since the products of the slopes of line PQ and RS is -1, the line PQ is perpendicular to the line RS
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