a. The sum of three numbers is 81. The second number is twice the first. given that the third number is 6 more than the second, find the numbers.
b. Give me the points P(3, 5) and Q(-5, 7) on the Cartesian plane such that R (x, y) is the midpoint of PQ, find the equation of the line that passes through R and perpendicular to line PQ.
Explanation
a. Let the numbers be p, r and q respectively
Given:
p + r + q = 81 --- (i)
r = 2p --- (ii)
q = r + 6 --- (iii)
Substitute (2p) for r in equation (iii)
q = 2p + 6 --- (iv)
Substitute (2p) for r and (2p + 6) for q in equation (i)
⇒ p + (2p) + (2p + 6) = 81
⇒ 5p + 6 = 81
⇒ 5p = 81 - 6
⇒ 5p = 75
p = \(\frac{75}{5}\) = 15.
From equation (ii)
⇒ r = 2(15) = 30
From equation (iv)
⇒ q = 2(15) + 6
⇒ q = 30 + 6 = 36
∴ The numbers are 15, 30 and 36
b. \(m_1 = \frac{ y_2 - y_1}{ x_2 - x_1}\)
Given points are P( 3, 5) and Q( -5, 7) = P(\( x_1 , y_1)\) , Q(\( x_2 , y_2)\)
the slope of line PQ = \((\frac{ 7 - 5}{ -5 - 3}) = \frac{2}{-8} = -\frac{1}{4}\)
Given that R( x, y) is the midpoint of the line formed by the points P( 3, 5) and Q( -5, 7)
R(x, y) = \(\frac{3 + (-5)}{2}, \frac{5 + 7}{2} = \frac{ 3 - 5}{2} , \frac{ 5+7}{2}\)
= -\(\frac{2}{2} , \frac{12}{2}\) = (-1, 6)
but the line passes through the point R( -1, 6) and is perpendicular to LINE PQ
\(m_2 = \frac{1}{m_1} = -\frac{1}{\frac{-1}{4}}\) = 4
Equation of the line can be obtained using \(y - y_1 = m_2( x - x_1)\)
y - 6 = 4( x - (-1)) = y - 6 = 4( x + 1 )
y - 6 = 4x + 4
y = 4x + 4 + 6
Therefore, y = 4x + 10